#include<iostream>
using namespace std;
int main()
{
    int N, M, Y;
    while (cin >> N >> M >> Y)
    {
        int f = 0,t=-1;
        for (int i = 0; i < M; i++)
        {
            int x = i;
            long long y = x;
            for (int j = 2; j < N + 1; j++)
            {
                y *= x;
                y %= M;
            }
            if (y== Y)
            {
                if (f > 0)
                    cout << ' ';
                cout << x ;
                f ++;
            }
        }
        if (f == 0)
            cout << t;
        cout << endl;
    }
    return 0;
}

why it can get accept,but this one get WA

#include<iostream>
using namespace std;
int main()
{
    int N, M, Y;
    while (cin >> N >> M >> Y)
    {
        int f = 0,t=-1;
        for (int i = 0; i < M; i++)
        {
            int x = i;
            long long y = x;
            for (int j = 2; j < N + 1; j++)
            {
                y *= x;
            }
            if (y%M== Y)
            {
                if (f > 0)
                    cout << ' ';
                cout << x ;
                f ++;
            }
        }
        if (f == 0)
            cout << t;
        cout << endl;
    }
    return 0;
}

I have a hard time submitting my first piece of code. Compilation error appeared and I have no idea why. The following is my code.

import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

/**
 * Created by Tanya on 2016/9/5.
 */
public class ReverseRoot2
{
    Scanner in;
    PrintWriter out;
    public void work()throws Exception
    {
        List<Double> numlist = new ArrayList<Double>();
        in = new Scanner(System.in);
        String line = "";
        while(in.hasNextLine() && !(line = in.nextLine()).equals(""))
        {
            String[] strnums = line.split(" ");
            for(String strnum : strnums)
            {
                if(strnum.matches("[0-9]+\\.*[0-9]*"))
                {
                    numlist.add(Double.parseDouble(strnum));
                }
            }
        }
        out = new PrintWriter(System.out);
        int len = numlist.size();
        for(int i = len - 1;i>=0;i--)
        {
            out.print(Math.sqrt(numlist.get(i).doubleValue()) + "\t");
        }
        out.flush();
    }
    static public void main(String[] args)throws Exception
    {
        ReverseRoot2 rr = new ReverseRoot2();
        rr.work();
    }
}

my code:

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <conio.h>
int SL (int n)
{
  int q=n/1000, i, s=0;
  while (q>10)
  {
    s+=q%10;
    q/=10;
  }
  s+=q;
  return s;
}
int SR (int n)
{
  int s=0,q=n%1000;
  while (q>10)
  {
    s+=q%10;
    q/=10;
  }
  s+=q;
  return s;
}
void main()
{
  int S=0,T=0,N;
  scanf ("%d", &N);
  if ((SR(N+1)==SL(N+1))||(SR(N-1)==SL(N-1)))
    printf ("Yes");
  else printf ("No");
}

#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;

int main() {
    long long a[100000] = {0};
    int i = 0;
    while (cin >> a[i]) {
        i++;
    }
    for (int j = i-1; j >= 0; j--) {
        cout << fixed << setprecision(4) << sqrt(a[j]) << endl;
    }
}

#include<iostream>
#include <cmath>
#include <climits>
#include <deque>

using namespace std;

int v[1001][1001];
float table[1002][1002];

int metro(int n,int m,int k){
    deque<int> qi,qj;
    qi.push_back(1);
    qj.push_back(1);
    table[1][1] = 0;
    while(!qi.empty()){
        int i = qi.front(),j = qj.front();
        qi.pop_front();
        qj.pop_front();
        if(i <= n && table[i+1][j] > table[i][j]+100){
            qi.push_back(i+1);
            qj.push_back(j);
            table[i+1][j] = table[i][j]+100;
        }
        if(j <= m && table[i][j+1] > table[i][j]+100){
            qi.push_back(i);
            qj.push_back(j+1);
            table[i][j+1] = table[i][j]+100;
        }
        if(v[i][j] == 1){
            if(i <= n && j <= m && table[i+1][j+1] > table[i][j] + 141.4){
                qi.push_back(i+1);
                qj.push_back(j+1);
                table[i+1][j+1] = table[i][j]+(141.4);
            }
        }
    }
    return ceil(table[n+1][m+1]);
}

int main(){
    int n = 0,m = 0,k = 0;
    cin>>n>>m;
    for(int i = 0;i <= n+1;++i){
        for(int j = 0;j <= m+1;++j){
            v[i][j] = 0;
            table[i][j] = INT_MAX;
        }
    }
    cin>>k;
    for(int i = 0;i < k;++i){
        int x = 0,y = 0;
        cin>>x>>y;
        v[x][y] = 1;
    }
    cout<<metro(n,m,k)<<endl;
    return 0;
}

Edited by author 29.08.2016 16:40

Edited by author 29.08.2016 16:40

Program Pr1;
var i:longint;
    A:real;
    k,n:longint;
begin
readln(n,k);
for i:= n downto k do begin
A:=sqrt(i);
writeln(A:0:4)
end;
end.

What is the test #85 ?

Edited by author 01.09.2016 22:25

Короче, баг был в коде for (int i = 1; i <= nmax; i++) adress[i] = NULL; - знак должен быть СТРОГИЙ!

Edited by author 26.08.2016 20:54

#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9+7, BLOCK = 320;

int n, m, a[100001], p[100001], idx[100001], order[100001], lb[100001], rb[100001]; // lb and rb are indexed by bfs order, not vertex
bool mark[100001];
vector<int> g[100001];

struct coolsqrt { // sqrt decomposition
    int f[100001], t[100001];
    void init() { for(int i = 1; i <= n; i++) f[i] = a[order[i]]; }
    void update(int l, int r, int v) {
        for(int i = l; i % BLOCK; i++) {
            f[i] = (f[i]+v)%MOD;
            if(i == r) return;
        }
        int block = l/BLOCK + 1;
        while(block < r/BLOCK) t[block++] += v, t[block] %= MOD;
        for(int i = BLOCK*(r/BLOCK); i <= r; i++) f[i]  = (f[i]+v)%MOD;
    }
    int query(int i) { return (f[i] + t[i/BLOCK])%MOD; }
    void putin() { for(int i = 1; i <= n; i++) cout << query(i) << ' '; cout << endl; }
} coolsqrt;

int main() {
    cin >> n;
    for(int i = 1; i <= n; i++) cin >> a[i];
    for(int i = 1,u,v; i < n; i++) {
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    queue<int> q;
    q.push(1);
    for(int i = 1, u = q.front(); i <= n; i++, q.pop(), u = q.front()) {
        mark[u] = true;
        idx[u] = i;
        order[i] = u;
        for(int v: g[u]) if(!mark[v]) q.push(v), p[v] = u;
    }
    assert(q.empty());
    for(int i = 1; i <= n; i++) if(p[order[i]]) {
        if(!lb[p[order[i]]]) lb[p[order[i]]] = i;
        rb[p[order[i]]] = i;
    }
    coolsqrt.init();
    cin >> m;
    int inst,v;
    while(m--) {
        cin >> inst >> v;
        if(inst == 1) {
            coolsqrt.update(idx[p[v]], idx[p[v]], coolsqrt.query(idx[v]));
            coolsqrt.update(lb[v], rb[v], coolsqrt.query(idx[v]));
            // coolsqrt.putin();
        } else {
            cout << coolsqrt.query(idx[v]) << endl;
        }
    }
    // for(int i = 1; i <= n; i++) cout << order[i] << ' ';
    // cout << endl;
    // for(int i = 1; i <= n; i++) cout << lb[i] << ' ' << rb[i] << endl;
}

Edited by author 01.09.2016 20:03

Edited by author 01.09.2016 20:03

Input can be such as this one:
14000( more than 225) (This problem can be on Pascal where there is byte type)
15000 14999 14888 14777 14666 .... 12313

So you should to change type for variable storing a number of tests)

Edited by author 06.04.2016 02:16

A = n * n;
B = n * n + n + 1;
C = n * n + 1;
A ^ 2 + B ^ 2
n^4 + (n ^ 2 + n + 1) * (n ^ 2 + n + 1)
n^4 + n ^ 4 + 2 * n^3 + 3*n^2 + 2*n + 1
2*n^4 + 2*n^3 + 2*n^2 + 2 * n + n^2 + 1
2 * n*(n ^ 3 + n ^ 2 + n + 1) + n^2 + 1
2 * n ( n(n ^ 2 + 1) + (n ^ 2 + 1)) + n^2 + 1
2 * n((n^2 + 1)(n + 1)) + (n ^ 2 + 1)
(n ^ 2 + 1)(2 * n * (n + 1) + 1)

If the input is P then the output will be 1 if P can be expressed as 1+(n(n-1)/2)

=> 1+(n(n-1)/2) = p
=> 2+n(n-1) = 2p
=> n(n-1) = 2p-2

To find such n I used floor and ceil of sqrt(2p-2)

What's wrong with my logic?

#include <bits/stdc++.h>

using namespace std;

int main(){
    double n, k, a, b, root;
    vector<int> v;
    cin >> k;
    while(k--){
        cin >> n;
        root = sqrt((2*n) - 2);
        a = ceil(root);
        b = floor(root);
        if(a == root && b == root){
            v.push_back(0);
        }else{
            if(a * b == (2*n) - 2) v.push_back(1);
            else v.push_back(0);
        }
    }
    for (int i = 0; i < v.size(); ++i){
        if(i == v.size() - 1) cout << v[i];
        else cout << v[i] << " ";
    }
    return 0;
}

Edited by author 30.08.2016 20:42

I am always wrong.

I've just brute-forced answers for N = 2...500 and found the trick.
But how to prove it is always correct or, at least, for the given binary sequences?

What is test 8?
I tested my program for 9<=n<=200 brute force and my answers is right.
For 1<=n<=8 output is:
i : NO when i<=n-1
n: aaaaaaaa (n times)

The test is "abcdeedcbaabcdeedcba"(without quotes).
The answer should be ""(empty string).

hey if one else to know print the anser plese say me..in the test #4 i have a empty string, but the judge say me "wa" in test #4.... why??? help me please.

#include <bits/stdc++.h>
using namespace std;

int main(){
    string cad,res;
             cin>>cad;
             res = "";
    int tam = cad.length();
    int i,j;
        i=0; j=1;

    int boolean[tam]={0};

    while(j<tam){
        if(cad[i]==cad[j]){
            boolean[i] = 1;
            boolean[j] = 1;

            if(i-1>=0){
                i-=1;
                j+=1;
            }else{
                i=j+1;
                j+=2;
            }
        }else{
            i=j;
            j+=1;
        }
    }

    for(int i=0;i<tam;i++){
         if(boolean[i]==0)
             res+=cad[i];
     }
     cout<<res;
    return 0;
}

My algo generate array like [1, -1, 2, -2, 3, -3 ...,0], but i got WA27
Any suggetion's?
(Bad english)

Edited by author 25.08.2016 23:47

1. Answer always exists.
2. Use DFS for all unmarked vertices and mark them: if current vertex is marked as x, all next must be marked as (x + 1) % 2. (0->1, 1->0)
3. Then just print all vertices marked by 0 (or 1).