GCC（大数取余）

题目描述

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. 

输入

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. 

0 < T <= 20
0 <= n < 10¹⁰⁰ (without leading zero) 
0 < m < 1000000 

输出

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

示例输入

1 10 861017 

示例输出

593846

解题思路：1,对于n!,当n>=m时，恒有n!%m==0。

  2,对于n!,n!%m=(n*(n-1)%m)*(n-1)*...2*1%m=(n*(n-1)*(n-2)%m)*(n-3)*...2*1=...，即对于阶乘取余，对于阶乘的每次积取余与总乘积的取余值相同，这点非常重要，需要重点理解并记住。

#include <stdio.h>#include <string.h>int main(){char s1[101],s2[7];int t,n,m,i,len1,len2,sum;long long int temp;scanf("%d",&t);while(t--){scanf("%s %s",s1,s2);len1=strlen(s1);len2=strlen(s2);for(i=m=0;i<len2;i++)m=m*10+s2[i]-'0';if(len1>len2 || (len1==len2 && strcmp(s1,s2)>=0))n=m-1;else{for(i=n=0;i<len1;i++)n=n*10+s1[i]-'0';}//printf("n=%d,m=%d\n",n,m);for(sum=i=0,temp=1;i<=n;i++){temp=temp*(i==0?1:i)%m;sum=(sum+temp)%m;}printf("%d\n",sum);}return 0;}