GCC + 数论

最新推荐文章于 2019-03-15 18:22:06 发布

李逝水

最新推荐文章于 2019-03-15 18:22:06 发布

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分类专栏：数论

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本文探讨了在GCC环境下解决阶乘累加求余问题的算法实现，涉及字符串解析、大数运算及取模操作。

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GCC

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3142 Accepted Submission(s): 986

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

593846
 
/*思路：如果n<=m，则计算(1!+2!+...+n!)%m
	而(1!+2!+...+n!)可以化解为1+1*(1+2*(1+3*(1+4*(1+5(.....(n-1)*(n)))))) ;
	返回n>m则只要计算(1!+2!+....+m!)%m同样可以求解出结果
*/
#include<iostream>
using namespace std ;
int main(void)
{
	int t , m ;
	char n[110] ;
	cin>> t;
	while(t--)
	{
		cin>> n >> m ;
		int i ;
		__int64 x  , sum = 0;
		for( i = 0 ; i < strlen(n) ; i ++){
			sum =sum*10+(n[i]-'0') ;
			if(sum >= m)
				break;
		}
		if(sum>=m){
			x = m - 1;
			for( i = x-1 ; i>=1 ; i --)
				x = ((( x + 1 )%m) * i%m)%m;
		}	
		else
		{
			x = sum ;
			for( i = x-1 ; i>=1 ; i --)
				x = ((( x + 1 )%m) * i%m)%m;
		}
		printf("%I64d\n",(x+1)%m);
	}
	return 0;
}