hdu - GCC - 取模与同余 - 数论

最新推荐文章于 2024-02-02 17:31:16 发布

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最新推荐文章于 2024-02-02 17:31:16 发布

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分类专栏： C/C++ 数论文章标签： GCC HDU 取模与同余

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本文介绍了一个基于GCC编译器的数学运算挑战问题，探讨了如何计算从0到n的阶乘之和并对m取模的问题。通过分析阶乘特性与取模运算之间的关系，提出了一种高效的解决方案。

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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

题解：n很大，int64也存不下，而据取模与同余，若n>m；(n>m)的部分将被m取余为0；所以只要0-m即可。

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;
const int N=1e6;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        char a[105];
        long long m;
        scanf("%s%lld",a,&m);
        int len=strlen(a);
        int i;
        long long n=0;
        for(i=0;i<len;i++)
        {
            n=n*10+(a[i]-'0');
            if(n>=m)
                break;
        }
        long long k=1,sum=1;
        if(n)
        {
            for(i=1;i<=m&&i<=n;i++)
            {
                k=(k*i)%m;
                sum=(sum+k)%m;
            }
        }
        printf("%lld\n",sum%m);
    }
    return 0;
}

备注：开始的时候答案报错，原因在36行，我没有对sum模m；因为当n=0，m=1时，sum=0而不是1。如何正整数对0取模是0。