SDUT 1919 规律，思维

最新推荐文章于 2022-11-20 21:25:30 发布

HAI__嗨I起来

最新推荐文章于 2022-11-20 21:25:30 发布

阅读量623

点赞数 1

CC 4.0 BY-SA版权

分类专栏：技巧 ----数学知识------ 大数

本文链接：https://blog.youkuaiyun.com/became_a_wolf/article/details/49454805

技巧同时被 3 个专栏收录

40 篇文章

订阅专栏

----数学知识------

34 篇文章

订阅专栏

大数

4 篇文章

订阅专栏

本文介绍了一个基于GCC的程序设计竞赛题目，任务是实现一个特殊的数学运算：计算从0!到n!的累加和并对m取模。文章提供了示例输入输出及程序代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

GCC

Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^

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 < 10 ¹⁰⁰(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

提示

来源

#include<bits/stdc++.h>
using namespace std;
long long int a[1000100];
int main()
{
    long long int n;
    int T;
    int m;
    char s[101];
    while(~scanf("%d",&T))
    {
        while(T--)
        {
            scanf("%lld%d",&n,&m);
            a[0]=1;
            int num=0;
            for(int i=1; i<=n; i++)
            {
                num++;
                a[i]=a[i-1]*i;
                a[i]%=m;
                if(a[i]==0)   //当i很大时候会整除m
                    break;
            }
            long long sum=0;
            for(int i=0; i<=num; i++)
            {
                sum+=a[i];
                sum%=m;
            }
            printf("%d\n",sum);
        }
    }
}
//1491

山东理工大学第三届ACM程序设计竞赛