Another Sum Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1477 Accepted Submission(s): 407

Problem Description

FunnyAC likes mathematics very much. He thinks mathematics is very funny and beautiful.When he solved a math problem he would be very happy just like getting accepted in ACM.Recently, he find a very strange problem.Everyone know that the sum of sequence from 1 to n is n*(n + 1)/2. But now if we create a sequence which consists of the sum of sequence from 1 to n. The new sequence is 1, 1+ 2, 1+2+3, .... 1+2+...+n. Now the problem is that what is the sum of the sequence from1 to 1+2+...+n .Is it very simple? I think you can solve it. Good luck!

Input

The first line contain an integer T .Then T cases followed. Each case contain an integer n (1 <= n <= 10000000).

Output

For each case,output the sum of first n items in the new sequence. Because the sum is very larger, so output sum % 20090524.

Sample Input

Sample Output

1
2600
30856

中間過程超了範圍..

#include<cstdio>
#include<iostream>
#define mod 20090524
using namespace std;
int main()
{
 int T=0;
 scanf("%d",&T);
 while(T--)
 {
  __int64 n=0;
  __int64 s[2]={2,3};
  scanf("%I64d",&n);
     __int64 a=n,b=n+1,c=n+2;
   for(int i=0;i<2;i++)
   {
    if(a%s[i]==0)
    {
     a/=s[i];
   }
   else if(b%s[i]==0)
   {
    b/=s[i];
   }
   else if(c%s[i]==0)
   {
    c/=s[i];
   }
   }
   __int64 sum=(((a%mod)*(b%mod))%mod*(c%mod))%mod;
     printf("%I64d\n",sum);
 }
 return 0;
}