ACM-ICPC An Olympian Math Problem（找规律）

最新推荐文章于 2021-08-22 07:42:46 发布

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本文介绍了一道关于阶乘和模运算的数学竞赛题目，通过观察规律简化计算过程，最终得出简便的解答方法。

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Alice, a student of grade 6, is thinking about an Olympian Math problem, but she feels so despair that she cries. And her classmate, Bob, has no idea about the problem. Thus he wants you to help him. The problem is:

We denote k!:

k! = 1 × 2 × ⋯ × (k − 1) × k

We denote S:

S = 1 × 1! + 2 × 2! + ⋯ + (n − 1) × (n − 1)!

Then S module n is

You are given an integer n.
You have to calculate S modulo n.

Input

The first line contains an integer T (T ≤ 1000), denoting the number of test cases. For each test case, there is a line which has an integer n.
It is guaranteed that 2 ≤ n ≤ 10^18 .

Output

For each test case, print an integer S modulo n.

Hint

The first test is: S = 1 × 1! = 1, and 1 modulo 2 is 1.

The second test is: S = 1 × 1! + 2 × 2! = 5 , and 5 modulo 3is 2.
样例输入复制
2
2
3
样例输出复制
1
2
打表找规律，可以简单列举前20个较小的数发现结果都是n-1。
打表代码：

#include<bits/stdc++.h>
#define LL long long
using namespace std;
LL del(LL a)
{
    LL temp=1;
    for(LL i=1;i<=a;i++)
        temp*=i;
    return temp;
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        LL n;scanf("%lld",&n);
        LL sum=0,k;
        for(LL i=1;i<n;i++)
        {
            sum+=i*del(i);
        }
        cout<<sum%n<<endl;
    }
    return 0;
}

AC代码：

#include<bits/stdc++.h>
#define LL long long
using namespace std;
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        LL n;scanf("%lld",&n);
        cout<<n-1<<endl;
    }
    return 0;
}