I have a program that finds, for all integers less than or equal to the input, numbers that can be represented as the sum of two cubes, twice, aka the Ramanujan's number problem.
I have written this in Java and Rust, however, it runs more than twice as slow in Rust as compared to Java.
Is there anything I can do to make it perform better, or otherwise improve it?
Rust code:
use num_integer::Roots;
fn main() {
let v = 984067;
// let v = 87539319;
for i in 1..=v {
ramanujan(i)
}
}
fn ramanujan(m: i32) {
let maxcube = m.cbrt();
let mut res1 = 0;
let mut res2 = 0;
let mut _res3 = 0;
let mut _res4 = 0;
for i in 1..=maxcube {
for j in 1..=maxcube {
if i * i * i + j * j * j == m {
res1 = i;
res2 = j;
break;
}
}
}
for k in 1..=maxcube {
for l in 1..=maxcube {
if k == res1 || k == res2 || l == res1 || l == res2 {
continue;
}
if k * k * k + l * l * l == m {
_res3 = k;
_res4 = l;
break;
}
}
}
// if ((res1 * res1 * res1) + (res2 * res2 * res2) == m) && ((res3 * res3 * res3) + (res4 * res4 * res4) == m) {
// println!("{} is representable as the sums of two different sets of two cubes!\nThese values are {}, {}, and {}, {}.", m, res1, res2, res3, res4);
// }
}
Java code:
public class Ramun {
public static void main(String[] args) {
int v = 984067;
// int v = 87539319;
for (int i = 1; i <= v; i++) {
ramanujan(i);
}
}
public static void ramanujan(int m) {
int maxcube = (int) Math.round(Math.cbrt(m));
int res1 = 0, res2 = 0, res3 = 0, res4 = 0;
for (int i = 1; i <= maxcube; i++) {
for (int j = 1; j <= maxcube; j++) {
if (((i * i * i) + (j * j * j)) == m) {
res1 = i;
res2 = j;
break;
}
}
}
for (int k = 1; k <= maxcube; k++) {
for (int l = 1; l <= maxcube; l++) {
if (k == res1 || k == res2 || l == res1 || l == res2)
continue;
if (((k * k * k) + (l * l * l)) == m) {
res3 = k;
res4 = l;
break;
}
}
}
// if (((res1 * res1 * res1) + (res2 * res2 * res2) == m) && ((res3 * res3 * res3) + (res4 * res4 * res4) == m)) {
// System.out.printf("%d is representable as the sums of two different sets of two cubes!%nThese values are %d, %d, and %d, %d.%n", m, res1, res2, res3, res4);
// }
}
}
Time output for both programs
The problem lies in RangeInclusive which can be expensive.
Here's a version avoiding it:
fn ramanujan(m: i32) {
let maxcube = m.cbrt() + 1; // we know it can't overflow
let mut res1 = 0;
let mut res2 = 0;
let mut res3 = 0;
let mut res4 = 0;
for i in 1..maxcube {
for j in 1..maxcube {
if i * i * i + j * j * j == m {
res1 = i;
res2 = j;
break;
}
}
}
for k in 1..maxcube {
for l in 1..maxcube {
if k == res1 || k == res2 || l == res1 || l == res2 {
continue;
}
if k * k * k + l * l * l == m {
res3 = k;
res4 = l;
break;
}
}
}
}
Result:
From: 0.01s user 0.00s system 0% cpu 17.993 total
To: 0.00s user 0.01s system 0% cpu 3.494 total
I added a comment to #45222 to draw attention to this issue.
Looks like for_each() allows better performance too (as for loops are more natural and should have the same performance, it should be considered as a bug):
fn ramanujan(m: i32) {
let maxcube = m.cbrt();
let mut res1 = 0;
let mut res2 = 0;
let mut res3 = 0;
let mut res4 = 0;
(1..=maxcube).for_each(|i| {
(1..=maxcube).try_for_each(|j| {
if i * i * i + j * j * j == m {
res1 = i;
res2 = j;
ControlFlow::Break(())
} else {
ControlFlow::Continue(())
}
});
});
(1..=maxcube).for_each(|k| {
(1..=maxcube).try_for_each(|l| {
if k != res1 && k != res2 && l != res1 && l != res2 && k * k * k + l * l * l == m {
res3 = k;
res4 = l;
ControlFlow::Break(())
} else {
ControlFlow::Continue(())
}
});
});
}
0.00s user 0.01s system 0% cpu 4.029 total
The code shows:
java.util.IllegalFormatConversionException: d != java.lang.String at
java.util.Formatter$FormatSpecifier.failConversion(Formatter.java:4302)
at
java.util.Formatter$FormatSpecifier.printInteger(Formatter.java:2793)
at java.util.Formatter$FormatSpecifier.print(Formatter.java:2747) a
public class addiePorterMod10Sieve {
void sieveOfEratosthenes(int n) {
boolean prime[] = new boolean[n + 1];
for (int i = 0; i < n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
for (int i = 2; i <= n; i++) {
if (prime[i] == true)
System.out.printf(i + "%-1s %-15n", " ");
}
}
public static void main(String args[]) {
int n = 1000;
addiePorterMod10Sieve g = new addiePorterMod10Sieve();
g.sieveOfEratosthenes(n);
}
}
Not very familiar with formatter myself, but the below workaround should achieve what you need as ouput :
int linecount = 0;
for(int i = 2; i <= n; i++)
{
if(prime[i] == true) {
// System.out.printf(i + "%-1s %-15d", " ");
linecount++;
System.out.print(i + " ");
if (linecount == 15) {
linecount =0;
System.out.println();
}
}
}
The error derives from %'s matching with a parameter.
int p = 0;
String nl = "\r\n";
for (int i = 2; i <= n; i++) {
if (prime[i]) {
++p;
System.out.printf("%-15d ", i);
if (p % 10 == 0) {
System.out.println();
}
//System.out.printf("%-15d%s", i, (p % 10 == 0 ? nl : " "));
}
}
Now %n would indeed yield a newline ("\r\n" on Windows, "\n" on Linux) with a flush of the line. However you must place it in the format string,
My out-commented alternative misses immediate flushing to the console.
I work with a Codility problem provided below,
The Fibonacci sequence is defined using the following recursive formula:
F(0) = 0
F(1) = 1
F(M) = F(M - 1) + F(M - 2) if M >= 2
A small frog wants to get to the other side of a river. The frog is initially located at one bank of the river (position −1) and wants to get to the other bank (position N). The frog can jump over any distance F(K), where F(K) is the K-th Fibonacci number. Luckily, there are many leaves on the river, and the frog can jump between the leaves, but only in the direction of the bank at position N.
The leaves on the river are represented in an array A consisting of N integers. Consecutive elements of array A represent consecutive positions from 0 to N − 1 on the river. Array A contains only 0s and/or 1s:
0 represents a position without a leaf;
1 represents a position containing a leaf.
The goal is to count the minimum number of jumps in which the frog can get to the other side of the river (from position −1 to position N). The frog can jump between positions −1 and N (the banks of the river) and every position containing a leaf.
For example, consider array A such that:
A[0] = 0
A[1] = 0
A[2] = 0
A[3] = 1
A[4] = 1
A[5] = 0
A[6] = 1
A[7] = 0
A[8] = 0
A[9] = 0
A[10] = 0
The frog can make three jumps of length F(5) = 5, F(3) = 2 and F(5) = 5.
Write a function:
class Solution { public int solution(int[] A); }
that, given an array A consisting of N integers, returns the minimum number of jumps by which the frog can get to the other side of the river. If the frog cannot reach the other side of the river, the function should return −1.
For example, given:
A[0] = 0
A[1] = 0
A[2] = 0
A[3] = 1
A[4] = 1
A[5] = 0
A[6] = 1
A[7] = 0
A[8] = 0
A[9] = 0
A[10] = 0
the function should return 3, as explained above.
Assume that:
N is an integer within the range [0..100,000];
each element of array A is an integer that can have one of the following values: 0, 1.
Complexity:
expected worst-case time complexity is O(N*log(N));
expected worst-case space complexity is O(N) (not counting the storage required for input arguments).
I wrote the following solution,
class Solution {
private class Jump {
int position;
int number;
public int getPosition() {
return position;
}
public int getNumber() {
return number;
}
public Jump(int pos, int number) {
this.position = pos;
this.number = number;
}
}
public int solution(int[] A) {
int N = A.length;
List<Integer> fibs = getFibonacciNumbers(N + 1);
Stack<Jump> jumps = new Stack<>();
jumps.push(new Jump(-1, 0));
boolean[] visited = new boolean[N];
while (!jumps.isEmpty()) {
Jump jump = jumps.pop();
int position = jump.getPosition();
int number = jump.getNumber();
for (int fib : fibs) {
if (position + fib > N) {
break;
} else if (position + fib == N) {
return number + 1;
} else if (!visited[position + fib] && A[position + fib] == 1) {
visited[position + fib] = true;
jumps.add(new Jump(position + fib, number + 1));
}
}
}
return -1;
}
private List<Integer> getFibonacciNumbers(int N) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < 2; i++) {
list.add(i);
}
int i = 2;
while (list.get(list.size() - 1) <= N) {
list.add(i, (list.get(i - 1) + list.get(i - 2)));
i++;
}
for (i = 0; i < 2; i++) {
list.remove(i);
}
return list;
}
public static void main(String[] args) {
int[] A = new int[11];
A[0] = 0;
A[1] = 0;
A[2] = 0;
A[3] = 1;
A[4] = 1;
A[5] = 0;
A[6] = 1;
A[7] = 0;
A[8] = 0;
A[9] = 0;
A[10] = 0;
System.out.println(solution(A));
}
}
However, while the correctness seems good, the performance is not high enough. Is there a bug in the code and how do I improve the performance?
Got 100% with simple BFS:
public class Jump {
int pos;
int move;
public Jump(int pos, int move) {
this.pos = pos;
this.move = move;
}
}
public int solution(int[] A) {
int n = A.length;
List < Integer > fibs = fibArray(n + 1);
Queue < Jump > positions = new LinkedList < Jump > ();
boolean[] visited = new boolean[n + 1];
if (A.length <= 2)
return 1;
for (int i = 0; i < fibs.size(); i++) {
int initPos = fibs.get(i) - 1;
if (A[initPos] == 0)
continue;
positions.add(new Jump(initPos, 1));
visited[initPos] = true;
}
while (!positions.isEmpty()) {
Jump jump = positions.remove();
for (int j = fibs.size() - 1; j >= 0; j--) {
int nextPos = jump.pos + fibs.get(j);
if (nextPos == n)
return jump.move + 1;
else if (nextPos < n && A[nextPos] == 1 && !visited[nextPos]) {
positions.add(new Jump(nextPos, jump.move + 1));
visited[nextPos] = true;
}
}
}
return -1;
}
private List < Integer > fibArray(int n) {
List < Integer > fibs = new ArrayList < > ();
fibs.add(1);
fibs.add(2);
while (fibs.get(fibs.size() - 1) + fibs.get(fibs.size() - 2) <= n) {
fibs.add(fibs.get(fibs.size() - 1) + fibs.get(fibs.size() - 2));
}
return fibs;
}
You can apply knapsack algorithms to solve this problem.
In my solution I precomputed fibonacci numbers. And applied knapsack algorithm to solve it. It contains duplicate code, did not have much time to refactor it. Online ide with the same code is in repl
import java.util.*;
class Main {
public static int solution(int[] A) {
int N = A.length;
int inf=1000000;
int[] fibs={1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025};
int[] moves = new int[N+1];
for(int i=0; i<=N; i++){
moves[i]=inf;
}
for(int i=0; i<fibs.length; i++){
if(fibs[i]-1<N && A[fibs[i]-1]==1){
moves[ fibs[i]-1 ] = 1;
}
if(fibs[i]-1==N){
moves[N] = 1;
}
}
for(int i=0; i<N; i++){
if(A[i]==1)
for(int j=0; j<fibs.length; j++){
if(i-fibs[j]>=0 && moves[i-fibs[j]]!=inf && moves[i]>moves[i-fibs[j]]+1){
moves[i]=moves[i-fibs[j]]+1;
}
}
System.out.println(i + " => " + moves[i]);
}
for(int i=N; i<=N; i++){
for(int j=0; j<fibs.length; j++){
if(i-fibs[j]>=0 && moves[i-fibs[j]]!=inf && moves[i]>moves[i-fibs[j]]+1){
moves[i]=moves[i-fibs[j]]+1;
}
}
System.out.println(i + " => " + moves[i]);
}
if(moves[N]==inf) return -1;
return moves[N];
}
public static void main(String[] args) {
int[] A = new int[4];
A[0] = 0;
A[1] = 0;
A[2] = 0;
A[3] = 0;
System.out.println(solution(A));
}
}
Javascript 100%
function solution(A) {
function fibonacciUntilNumber(n) {
const fib = [0,1];
while (true) {
let newFib = fib[fib.length - 1] + fib[fib.length - 2];
if (newFib > n) {
break;
}
fib.push(newFib);
}
return fib.slice(2);
}
A.push(1);
const fibSet = fibonacciUntilNumber(A.length);
if (fibSet.includes(A.length)) return 1;
const reachable = Array.from({length: A.length}, () => -1);
fibSet.forEach(jump => {
if (A[jump - 1] === 1) {
reachable[jump - 1] = 1;
}
})
for (let index = 0; index < A.length; index++) {
if (A[index] === 0 || reachable[index] > 0) {
continue;
}
let minValue = 100005;
for (let jump of fibSet) {
let previousIndex = index - jump;
if (previousIndex < 0) {
break;
}
if (reachable[previousIndex] > 0 && minValue > reachable[previousIndex]) {
minValue = reachable[previousIndex];
}
}
if (minValue !== 100005) {
reachable[index] = minValue + 1;
}
}
return reachable[A.length - 1];
}
Python 100% answer.
For me the easiest solution was to locate all leaves within one fib jump of -1. Then consider each of these leaves to be index[0] and find all jumps from there.
Each generation or jump is recorded in a set until a generation contains len(A) or no more jumps can be found.
def gen_fib(n):
fn = [0,1]
i = 2
s = 2
while s < n:
s = fn[i-2] + fn[i-1]
fn.append(s)
i+=1
return fn
def new_paths(A, n, last_pos, fn):
"""
Given an array A of len n.
From index last_pos which numbers in fn jump to a leaf?
returns list: set of indexes with leaves.
"""
paths = []
for f in fn:
new_pos = last_pos + f
if new_pos == n or (new_pos < n and A[new_pos]):
paths.append(new_pos)
return path
def solution(A):
n = len(A)
if n < 3:
return 1
# A.append(1) # mark final jump
fn = sorted(gen_fib(100000)[2:]) # Fib numbers with 0, 1, 1, 2.. clipped to just 1, 2..
# print(fn)
paths = set([-1]) # locate all the leaves that are one fib jump from the start position.
jump = 1
while True:
# Considering each of the previous jump positions - How many leaves from there are one fib jump away
paths = set([idx for pos in paths for idx in new_paths(A, n, pos, fn)])
# no new jumps means game over!
if not paths:
break
# If there was a result in the new jumps record that
if n in paths:
return jump
jump += 1
return -1
https://app.codility.com/demo/results/training4GQV8Y-9ES/
https://github.com/niall-oc/things/blob/master/codility/fib_frog.py
Got 100%- solution in C.
typedef struct state {
int pos;
int step;
}state;
int solution(int A[], int N) {
int f1 = 0;
int f2 = 1;
int count = 2;
// precalculating count of maximum possible fibonacci numbers to allocate array in next loop. since this is C language we do not have flexible dynamic structure as in C++
while(1)
{
int f3 = f2 + f1;
if(f3 > N)
break;
f1 = f2;
f2 = f3;
++count;
}
int fib[count+1];
fib[0] = 0;
fib[1] = 1;
int i = 2;
// calculating fibonacci numbers in array
while(1)
{
fib[i] = fib[i-1] + fib[i-2];
if(fib[i] > N)
break;
++i;
}
// reversing the fibonacci numbers because we need to create minumum jump counts with bigger jumps
for(int j = 0, k = count; j < count/2; j++,k--)
{
int t = fib[j];
fib[j] = fib[k];
fib[k] = t;
}
state q[N];
int front = 0 ;
int rear = 0;
q[0].pos = -1;
q[0].step = 0;
int que_s = 1;
while(que_s > 0)
{
state s = q[front];
front++;
que_s--;
for(int i = 0; i <= count; i++)
{
int nextpo = s.pos + fib[i];
if(nextpo == N)
{
return s.step+1;
}
else if(nextpo > N || nextpo < 0 || A[nextpo] == 0){
continue;
}
else
{
q[++rear].pos = nextpo;
q[rear].step = s.step + 1;
que_s++;
A[nextpo] = 0;
}
}
}
return -1;
}
//100% on codility Dynamic Programming Solution. https://app.codility.com/demo/results/training7WSQJW-WTX/
class Solution {
public int solution(int[] A) {
int n = A.length + 1;
int dp[] = new int[n];
for(int i=0;i<n;i++) {
dp[i] = -1;
}
int f[] = new int[100005];
f[0] = 1;
f[1] = 1;
for(int i=2;i<100005;i++) {
f[i] = f[i - 1] + f[i - 2];
}
for(int i=-1;i<n;i++) {
if(i == -1 || dp[i] > 0) {
for(int j=0;i+f[j] <n;j++) {
if(i + f[j] == n -1 || A[i+f[j]] == 1) {
if(i == -1) {
dp[i + f[j]] = 1;
} else if(dp[i + f[j]] == -1) {
dp[i + f[j]] = dp[i] + 1;
} else {
dp[i + f[j]] = Math.min(dp[i + f[j]], dp[i] + 1);
}
}
}
}
}
return dp[n - 1];
}
}
Ruby 100% solution
def solution(a)
f = 2.step.inject([1,2]) {|acc,e| acc[e] = acc[e-1] + acc[e-2]; break(acc) if acc[e] > a.size + 1;acc }.reverse
mins = []
(a.size + 1).times do |i|
next mins[i] = -1 if i < a.size && a[i] == 0
mins[i] = f.inject(nil) do |min, j|
k = i - j
next min if k < -1
break 1 if k == -1
next min if mins[k] < 0
[mins[k] + 1, min || Float::INFINITY].min
end || -1
end
mins[a.size]
end
I have translated the previous C solution to Java and find the performance is improved.
import java.util.*;
class Solution {
private static class State {
int pos;
int step;
public State(int pos, int step) {
this.pos = pos;
this.step = step;
}
}
public static int solution(int A[]) {
int N = A.length;
int f1 = 0;
int f2 = 1;
int count = 2;
while (true) {
int f3 = f2 + f1;
if (f3 > N) {
break;
}
f1 = f2;
f2 = f3;
++count;
}
int[] fib = new int[count + 1];
fib[0] = 0;
fib[1] = 1;
int i = 2;
while (true) {
fib[i] = fib[i - 1] + fib[i - 2];
if (fib[i] > N) {
break;
}
++i;
}
for (int j = 0, k = count; j < count / 2; j++, k--) {
int t = fib[j];
fib[j] = fib[k];
fib[k] = t;
}
State[] q = new State[N];
for (int j = 0; j < N; j++) {
q[j] = new State(-1,0);
}
int front = 0;
int rear = 0;
// q[0].pos = -1;
// q[0].step = 0;
int que_s = 1;
while (que_s > 0) {
State s = q[front];
front++;
que_s--;
for (i = 0; i <= count; i++) {
int nextpo = s.pos + fib[i];
if (nextpo == N) {
return s.step + 1;
}
//
else if (nextpo > N || nextpo < 0 || A[nextpo] == 0) {
continue;
}
//
else {
q[++rear].pos = nextpo;
q[rear].step = s.step + 1;
que_s++;
A[nextpo] = 0;
}
}
}
return -1;
}
}
JavaScript with 100%.
Inspired from here.
function solution(A) {
const createFibs = n => {
const fibs = Array(n + 2).fill(null)
fibs[1] = 1
for (let i = 2; i < n + 1; i++) {
fibs[i] = fibs[i - 1] + fibs[i - 2]
}
return fibs
}
const createJumps = (A, fibs) => {
const jumps = Array(A.length + 1).fill(null)
let prev = null
for (i = 2; i < fibs.length; i++) {
const j = -1 + fibs[i]
if (j > A.length) break
if (j === A.length || A[j] === 1) {
jumps[j] = 1
if (prev === null) prev = j
}
}
if (prev === null) {
jumps[A.length] = -1
return jumps
}
while (prev < A.length) {
for (let i = 2; i < fibs.length; i++) {
const j = prev + fibs[i]
if (j > A.length) break
if (j === A.length || A[j] === 1) {
const x = jumps[prev] + 1
const y = jumps[j]
jumps[j] = y === null ? x : Math.min(y, x)
}
}
prev++
while (prev < A.length) {
if (jumps[prev] !== null) break
prev++
}
}
if (jumps[A.length] === null) jumps[A.length] = -1
return jumps
}
const fibs = createFibs(26)
const jumps = createJumps(A, fibs)
return jumps[A.length]
}
const A = [0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0]
console.log(A)
const s = solution(A)
console.log(s)
You should use a QUEUE AND NOT A STACK. This is a form of breadth-first search and your code needs to visit nodes that were added first to the queue to get the minimum distance.
A stack uses the last-in, first-out mechanism to remove items while a queue uses the first-in, first-out mechanism.
I copied and pasted your exact code but used a queue instead of a stack and I got 100% on codility.
100% C++ solution
More answers in my github
Inspired from here
Solution1 : Bottom-Top, using Dynamic programming algorithm (storing calculated values in an array)
vector<int> getFibonacciArrayMax(int MaxNum) {
if (MaxNum == 0)
return vector<int>(1, 0);
vector<int> fib(2, 0);
fib[1] = 1;
for (int i = 2; fib[fib.size()-1] + fib[fib.size() - 2] <= MaxNum; i++)
fib.push_back(fib[i - 1] + fib[i - 2]);
return fib;
}
int solution(vector<int>& A) {
int N = A.size();
A.push_back(1);
N++;
vector<int> f = getFibonacciArrayMax(N);
const int oo = 1'000'000;
vector<int> moves(N, oo);
for (auto i : f)
if (i - 1 >= 0 && A[i-1])
moves[i-1] = 1;
for (int pos = 0; pos < N; pos++) {
if (A[pos] == 0)
continue;
for (int i = f.size()-1; i >= 0; i--) {
if (pos + f[i] < N && A[pos + f[i]]) {
moves[pos + f[i]] = min(moves[pos]+1, moves[pos + f[i]]);
}
}
}
if (moves[N - 1] != oo) {
return moves[N - 1];
}
return -1;
}
Solution2: Top-Bottom using set container:
#include <set>
int solution2(vector<int>& A) {
int N = A.size();
vector<int> fib = getFibonacciArrayMax(N);
set<int> positions;
positions.insert(N);
for (int jumps = 1; ; jumps++)
{
set<int> new_positions;
for (int pos : positions)
{
for (int f : fib)
{
// return jumps if we reach to the start point
if (pos - (f - 1) == 0)
return jumps;
int prev_pos = pos - f;
// we do not need to calculate bigger jumps.
if (prev_pos < 0)
break;
if (prev_pos < A.size() && A[prev_pos])
new_positions.insert(prev_pos);
}
}
if (new_positions.size() == 0)
return -1;
positions = new_positions;
}
return -1;
}
Below is a Archive PROBLEM from SPOJ. Sample testCase is passing, but I am getting W/A on submission. I am missing some testCase(testCases). Need help to figure out what case I am missing and/or what I am doing wrong here.
Ada the Ladybug is playing Game of Divisors against her friend Velvet Mite Vinit. The game has following rules. There is a pile of N stones between them. The player who's on move can pick at least 1 an at most σ(N) stones (where σ(N) stands for number of divisors of N). Obviously, N changes after each move. The one who won't get any stones (N == 0) loses.
As Ada the Ladybug is a lady, so she moves first. Can you decide who will be the winner? Assume that both players play optimally.
Input
The first line of input will contain 1 ≤ T ≤ 10^5, the number of test-cases.
The next T lines will contain 1 ≤ N ≤ 2*10^7, the number of stones which are initially in pile.
Output
Output the name of winner, so either "Ada" or "Vinit".
Sample Input:
8
1
3
5
6
11
1000001
1000000
29
Sample Output:
Ada
Vinit
Ada
Ada
Vinit
Vinit
Ada
Ada
CODE
import java.io.*;
public class Main
{
public static int max_size = 2 * (int)Math.pow(10,7) + 1;
//public static int max_size = 25;
//public static int max_size = 2 * (int)Math.pow(10,6) + 1;
public static boolean[] dp = new boolean[max_size];
public static int[] lastPrimeDivisor = new int[max_size];
public static int[] numOfDivisors = new int[max_size];
public static void main(String[] args) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
preprocess();
int t = Integer.parseInt(br.readLine());
while(t > 0)
{
int n = Integer.parseInt(br.readLine());
if(dp[n] == true)
System.out.println("Ada");
else
System.out.println("Vinit");
t--;
}
}
public static void markLastPrimeDivisor()
{
for(int i = 0 ; i < max_size ; i++)
{
lastPrimeDivisor[i] = 1;
}
for(int i = 2 ; i < max_size ; i += 2)
{
lastPrimeDivisor[i] = 2;
}
int o = (int)Math.sqrt(max_size);
for(int i = 3 ; i < max_size; i++)
{
if(lastPrimeDivisor[i] != 1)
{
continue;
}
lastPrimeDivisor[i] = i;
if(i <= o)
{
for(int j = i * i ; j < max_size ; j += 2 * i)
{
lastPrimeDivisor[j] = i;
}
}
}
/*for(int i = 1 ; i < max_size ; i++)
System.out.println("last prime of " + i + " is " + lastPrimeDivisor[i]);*/
}
public static void countDivisors(int num)
{
int original = num;
int result = 1;
int currDivisorCount = 1;
int currDivisor = lastPrimeDivisor[num];
int nextDivisor;
while(currDivisor != 1)
{
num = num / currDivisor;
nextDivisor = lastPrimeDivisor[num];
if(nextDivisor == currDivisor)
{
currDivisorCount++;
}
else
{
result = result * (currDivisorCount + 1);
currDivisorCount = 1;
currDivisor = nextDivisor;
}
}
if(num != 1)
{
result = result * (currDivisorCount + 1);
}
//System.out.println("result for num : " + original + ", " + result);
numOfDivisors[original] = result;
}
public static void countAllDivisors()
{
markLastPrimeDivisor();
for(int i = 2 ; i < max_size ; i++)
{
countDivisors(i);
//System.out.println("num of divisors of " + i + " = " + numOfDivisors[i]);
}
}
public static void preprocess()
{
countAllDivisors();
dp[0] = dp[1] = dp[2] = true;
for(int i = 3 ; i < max_size ; i++)
{
int flag = 0;
int limit = numOfDivisors[i];
//If for any i - j, we get false,for playing optimally
//the current opponent will choose to take j stones out of the
//pile as for i - j stones, the other player is not winning.
for(int j = 1 ; j <= limit; j++)
{
if(dp[i - j] == false)
{
dp[i] = true;
flag = 1;
break;
}
}
if(flag == 0)
dp[i] = false;
}
}
}
There is a subtle bug in your countDivisors() function. It assumes
that lastPrimeDivisor[num] – as the name indicates – returns the
largest prime factor of the given argument.
However, that is not the case. For example, lastPrimeDivisor[num] = 2
for all even numbers, or lastPrimeDivisor[7 * 89] = 7.
The reason is that in
public static void markLastPrimeDivisor()
{
// ...
for(int i = 3 ; i < max_size; i++)
{
// ...
if(i <= o)
{
for(int j = i * i ; j < max_size ; j += 2 * i)
{
lastPrimeDivisor[j] = i;
}
}
}
}
only array elements starting at i * i are updated.
So lastPrimeDivisor[num] is in fact some prime divisor of num, but not
necessarily the largest. As a consequence, numOfDivisors[55447] is computed
as 8 instead of the correct value 6.
Therefore in countDivisors(), the exponent of a prime factor in num
must be determined explicitly by repeated division.
Then you can use that the divisors function is multiplicative. This leads to
the following implementation:
public static void countAllDivisors() {
// Fill the `somePrimeDivisor` array:
computePrimeDivisors();
numOfDivisors[1] = 1;
for (int num = 2 ; num < max_size ; num++) {
int divisor = somePrimeDivisor[num];
if (divisor == num) {
// `num` is a prime
numOfDivisors[num] = 2;
} else {
int n = num / divisor;
int count = 1;
while (n % divisor == 0) {
count++;
n /= divisor;
}
// `divisor^count` contributes to `count + 1` in the number of divisors,
// now use multiplicative property:
numOfDivisors[num] = (count + 1) * numOfDivisors[n];
}
}
}
What am i doing wrong here?
Java code for computing prefix function. Two input are right but the last one is wrong.
Here's the pseudocode:
Java code:
class Main {
// compute prefix function
public static void main(String[] args) {
String p = "422213422153342";
String x = "ababbabbabbababbabb";
String y = "ababaca";
printOutput(p);
printOutput(y);
System.out.println();System.out.println();
System.out.println("the prefix func below is wrong. I am not sure why.");
System.out.print("answer should be: 0 0 1 2 0 1 2 0 1 2 0 1 2 3 4 5 6 7 8");
printOutput(x);
}
static void printOutput(String P){
System.out.println();System.out.println();
System.out.print("p[i]: ");
for(int i = 0; i < P.length(); i++)System.out.print(P.charAt(i) + " ");
System.out.println();
System.out.print("Pi[i]: ");
compute_prefix_func(P);
}
public static void compute_prefix_func(String P){
int m = P.length();
int pi[] = new int[m];
for(int i = 0; i < pi.length; i++){
pi[i] = 0;
}
pi[0] = 0;
int k = 0;
for(int q = 2; q < m; q++){
while(k > 0 && ( ((P.charAt(k) + "").equals(P.charAt(q) + "")) == false)){
k = pi[k];
}
if ((P.charAt(k) + "").equals(P.charAt(q) + "")){
k = k + 1;
}
pi[q] = k;
}
for(int i = 0; i < pi.length; i++){
System.out.print(pi[i] + " ");
}
}
}
Okay, let's start off by making the code much easier to read. This:
if ((P.charAt(k) + "").equals(P.charAt(q) + ""))
can be simplified to:
if (P.charAt(k) == P.charAt(q))
... and you've done that in multiple places.
Likewise here:
int pi[] = new int[m];
for(int i = 0; i < pi.length; i++){
pi[i] = 0;
}
pi[0] = 0;
... you don't need the explicit initialization. Variables are 0-initialized by default. (It's unclear why you're then setting pi[0] again, although I note that if P.length() is 0, this will throw an exception.)
Next is to remove the explicit comparison with false, instead just using ! so we have:
while(k > 0 && P.charAt(k) != P.charAt(q))
Finally, let's restructure the code a bit to make it easier to follow, use more conventional names, and change int pi[] to the more idiomatic int[] pi:
class Main {
public static void main(String[] args) {
String x = "ababbabbabbababbabb";
int[] prefix = computePrefix(x);
System.out.println("Prefix series for " + x);
for (int p : prefix) {
System.out.print(p + " ");
}
System.out.println();
}
public static int[] computePrefix(String input) {
int[] pi = new int[input.length()];
int k = 0;
for(int q = 2; q < input.length(); q++) {
while (k > 0 && input.charAt(k) != input.charAt(q)) {
k = pi[k];
}
if (input.charAt(k) == input.charAt(q)) {
k = k + 1;
}
pi[q] = k;
}
return pi;
}
}
That's now much easier to follow, IMO.
We can now look back to the pseudocode and see that it appears to be using 1-based indexing for both arrays and strings. That makes life slightly tricky. We could mimic that throughout the code, changing every array access and charAt call to just subtract 1.
(I've extracted the common subexpression of P[q] to a variable target within the loop.)
public static int[] computePrefix(String input) {
int[] pi = new int[input.length()];
int k = 0;
for (int q = 2; q <= input.length(); q++) {
char target = input.charAt(q - 1);
while (k > 0 && input.charAt(k + 1 - 1) != target) {
k = pi[k - 1];
}
if (input.charAt(k + 1 - 1) == target) {
k++;
}
pi[q - 1] = k;
}
return pi;
}
That now gives your desired results, but it's really ugly. We can shift q very easily, and remove the + 1 - 1 parts:
public static int[] computePrefix(String input) {
int[] pi = new int[input.length()];
int k = 0;
for (int q = 1; q < input.length(); q++) {
char target = input.charAt(q);
while (k > 0 && input.charAt(k) != target) {
k = pi[k - 1];
}
if (input.charAt(k) == target) {
k++;
}
pi[q] = k;
}
return pi;
}
It's still not entirely pleasant, but I think it's what you want. Make sure you understand why I had to make the changes I did.
public static int[] computePrefix(String input) {
int[] pi = new int[input.length()];
pi[0] = -1;
int k = -1;
for (int q = 1; q < input.length(); q++) {
char target = input.charAt(q);
while (k > 0 && input.charAt(k + 1) != target) {
k = pi[k];
}
if (input.charAt(k + 1) == target) {
k++;
}
pi[q] = k;
}
return pi;
}