杭电acm 3132GCC（简单数论）

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#杭电acm 3132GCC简单数论

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本文介绍了一个基于GCC编译器系统的编程挑战，任务是计算一系列阶乘之和再对某个数取余的操作。文章提供了问题描述、输入输出格式、样例及解析，并附带了解题思路与实现代码。

GCC

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 5140 Accepted Submission(s): 1716

Problem Description

The GNU Compiler Collection (usually shortened to GCC) is a compiler system produced by the GNU Project supporting various programming languages. But it doesn’t contains the math operator “!”.
In mathematics the symbol represents the factorial operation. The expression n! means "the product of the integers from 1 to n". For example, 4! (read four factorial) is 4 × 3 × 2 × 1 = 24. (0! is defined as 1, which is a neutral element in multiplication, not multiplied by anything.)
We want you to help us with this formation: (0! + 1! + 2! + 3! + 4! + ... + n!)%m

Input

The first line consists of an integer T, indicating the number of test cases.
Each test on a single consists of two integer n and m.

Output

Output the answer of (0! + 1! + 2! + 3! + 4! + ... + n!)%m.

Constrains
0 < T <= 20
0 <= n < 10^100 (without leading zero)
0 < m < 1000000

Sample Input

  
   1
10 861017

Sample Output

Source

2009 Asia Wuhan Regional Contest Online

想法：

n>m时，比m大的数乘起后对取余m后没影响

代码：

#include<stdio.h>
#include<string.h>
int main()
{
int i,j,n,m,cas,t;
long long mid,sum;
char s[1000];
scanf("%d",&cas);
while(cas--)
{
scanf("%s %d",&s,&m);
if(strlen(s)>=7)
t=m;
else sscanf(s,"%d",&t);
mid=1;
sum=1;
for(i=1;i<=t;i++)
{
mid=(mid*i)%m;
sum=(sum+mid)%m;
}
printf("%lld\n",sum%m);//这里一定一定要%m 否则会错
}
return 0;
}