HDU 2576 Another Sum Problem

Another Sum Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1800    Accepted Submission(s): 494

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

 

Source

HDU 2009-5 Programming Contest

 

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利用多项式求和公式1*2/2+2*3/2+3*4/2+……+n(n+1)/2=n(n+1)(n+2)/6

此外还要注意取余运算。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){int t,k,l;__int64 m,n;scanf("%d",&t);while(t--){scanf("%I64d",&n);m=(n*(n+1))%(20090524*6);m=(m*(n+2)/6)%20090524;printf("%I64d\n",m);}return 0;}