HDU 2576 数学

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

312456

 

Sample Output

1260030856
 
中间过程超了范围..
#include<cstdio>#include<iostream>#define mod 20090524using 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;}