import java.util.Scanner;

public class tests {


    public static void main(String[] args) {
             int i=0;

                Scanner scan = new Scanner(System.in);
                try{
                int n=scan.nextInt();
                if ((n>=1)&&(n<=10000)){
                    i=n *(n + 1) / 2;
                    System.out.print(i);}
                else
                if ((n<=1)&&(n>=-10000)){
                    i=-((-n) *(1 -n) / 2) + 1;
                    System.out.print(i);}
                }
                finally
                {
                    scan.close();
                    }
        }
}

try this:
45000270000405
answer:
9000054000081

or this:
907500003300000003
answer:
302500001100000001

Edited by author 10.10.2011 10:41

My solve in C++:
#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <stack>
#include <algorithm>
#include <bitset>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <iomanip>

#define F first
#define S second
#define ll long long
#define len length()
#define sqr(x) x*x
#define pb push_back
#define mp make_pair
#define sz(x) ((int) (x).size())
#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()
#define bp(x) __builtin_popcount(x)
#define INF numeric_limits<long long int>::max()
#define frp freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
#define forit(it, s) for(__typeof(s.begin()) it = s.begin(); it != s.end(); it++)

const int maxn = (int)1e6;
const int mod = (int)1e9 + 7;

using namespace std;

__int64 n,m,y;

int binpow(int a, int n){
    if (n == 0)
        return 1;
    if (n % 2 == 1)
        return (binpow (a, n-1)*a)%m;
    else{
        int b = (binpow(a, n/2));
        return (b*b)%m;
    }
}

bool ok;

main(){
      scanf("%I64d%I64d%I64d",&n,&m,&y);
      for(int i=0; i < m; i++){
          if(binpow(i,n)%m==y){
              cout<<i<<" ";
              ok=1;
          }
      }
      if(!ok)
          return cout<<"-1",0;

    return 0;
}

What it is?
  else if y mod 2=0 then BinPow:=sqr(BinPow(x,y div 2)) mod M
  else BinPow:=(sqr(BinPow(x,y div 2))*x) mod M;

In the end you write empty line. You must delete your last cout.

Error here: t=0. You wanted this:
#include<stdio.h>
int main()
{
    int n,i,j,c,t,T;
    scanf("%d",&T);
    while(T--)
    {
    scanf("%d",&n);
    t=0;
    for(j=2;;j++)
    {
        c=0;
        for(i=2;i<=j/2;i++)
        {
            if(j%i==0)
            {
                c++;
                break;
            }
        }
        if(c==0)
            t++;
        if(t==n)
           {
               printf("%d\n",j);
               break;
            }

    }
  }
  return 0;
}

But your algo will get TL on test:
15000
15000
....
15000
(15001 times)

correst your x and y to:

[code deleted]

Edited by author 21.11.2005 04:45

Edited by moderator 22.02.2006 00:49

Hi. I've also got AC with O(m*2^n), but it seems that it's a bit more fast. Here's an idea.
Each set of taps or a combination of sets has corresponding bit mask. We have an array int minPrice[2^N]. Obviously, index is a mask, element — min price for buying such combination of taps. Initially the only !=0 elements are those each corresponding to one of predefined M sets.
Then the DP comes. We go from 0-th element to the last (11...1 target). If i-th element !=0, we process it: a) if i corresponds to one of M predefined sets, we try to OR "i" with all younger mask with !=0 price. b) if i doesn't correspond to any predefined set, thus, it's a combination of some predefined sets, then we OR this "i" with younger predefined masks ONLY (not all younger "j" !=0).

Gold is first 4 teams

Hi,

The output of my application for your test is the same. But the judge tell me WA#8 :S.

Thanks, man, that helped me :)

Me too!Have you solved this problem?

I have this WA too.
Firstly, I used binary search to find the last element, which didn't fall.

And my mistake was:
if (currentSum == k)
 return med;

the corrent ones:
if (currentSum == k)
 return med-1;

And for case, where we haven't found exactly k sum:

..
}//end of binary search

int answerIndex = l;
int sum = getSum(left, lastAdded Index);
if (sum <= k){
    answerIndex = l - 1;
}
return answerIndex;


But now I have TL11

5 1 4
1 2 3 4 5
5
ans: 8