hdoj--3123--GCC（技巧阶乘取余）-优快云博客

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的和并对某个数取模。通过分析输入数字的大小，文章提供了一种有效的算法实现，能够处理大整数的情况。

GCC

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4473 Accepted Submission(s): 1472

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

Source

2009 Asia Wuhan Regional Contest Online

好吧，0！是1，我服了，，，，，

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define ll long long
#define N 1000010
using namespace std;
char s[110];
ll sum[N];
ll change(char *s)
{
	int l,i,ans=0;
	l=strlen(s);
	for(i=0;i<l;i++)
		ans=ans*10+(s[i]-'0');
	return ans; 
}
int main()
{
	int t;
	int n,m;
	int i,j,k,l;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s%d",s,&m);
		l=strlen(s);
		if(l<7)
		{
			n=change(s);
			if(n>m)
				n=m-1;
			sum[0]=1;
			int ans=0;
			for(i=1;i<=n;i++)
			{
				sum[i]=(sum[i-1]*i)%m;
				ans=(ans+sum[i])%m;
			}
			printf("%d\n",(ans+1)%m);
		}
		else
		{
			n=m-1;
			sum[0]=1;
			int ans=0;
			for(i=1;i<=n;i++)
			{
				sum[i]=(sum[i-1]*i)%m;
				ans=(ans+sum[i])%m;
			}
			printf("%d\n",(ans+1)%m);
		}
	}
	return 0;
}

转载于:https://www.cnblogs.com/playboy307/p/5273629.html