HDU 2576 Another Sum Problem

传送门：Another Sum Problem

Another Sum Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1227    Accepted Submission(s): 336

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

 

解题报告：

数论题：可以推公式。

公式如下：

假设x=sn则 sn-sn-1=n(n+1)/2; s1=1；求sn的通项公式  通过累加法  之后化简 可求得公式为Sn = n(n+1)(n+2)/6;对sn求余 我们可以分成2部分    即n(n+1)   和  (n+2)注意 本题中要保证n(n+1)(n+2)能被6整除  因为sn一定是个整数所以求余的时候要这样求余s1 = (n(n+1))% (20090524*6);余数里包含6是保证后面的式子可以被6整除。s2 = (s1*(n+2)/6)%20090524;

代码如下：

#include<stdio.h>#define MOD 20090524int main(){    int t;    long long n;    scanf("%d",&t);    while(t--){        scanf("%I64d",&n);  //注意，使用lld答案会错误        if(n==1)            printf("1\n");        else{            long long ans=(n*(n+1))%(MOD*6);            ans=(ans*(n+2)/6)%MOD;            printf("%lld\n",ans);        }    }    return 0;}