An Olympian Math Problem(c++)

原创于 2021-08-22 07:42:46 发布 · 278 阅读

0 ·

CC 4.0 BY-SA版权

2006-1223-诺

文章标签：

#c++ #算法

c++ 同时被 2 个专栏收录

9 篇文章

订阅专栏

计蒜客

6 篇文章

订阅专栏

本文解析了一个关于阶乘和求和的奥林匹克数学问题，指出关键是理解S=1!×(1+2+...+(n-1))，并揭示了计算S模n的解法：只需输出n-1。实例演示和代码实现帮助学生Bob快速掌握解题方法。

Alice, a student of grade 66, 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)×(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 nn.

It is guaranteed that 2≤n≤10^18.

Output

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

Hint

The first test is: S = 1*(1!)= 1;

S=1×1!=1, and 1 modulo 2 is 1.

The second test is: S = 1*1!+2*2!= 5;S=1×1!+2×2!=5 , and 5 modulo 3 is 2.

样例输入

2
2
3

样例输出

1
2

题目来源

ACM-ICPC 2018 南京赛区网络预赛

不要看着这题很恶心啊，其实找完规律之后你会发现很简单hhh

又双叒叕是作者的草图时间啊哈哈哈啊哈哈太草了

！是阶乘大家应该懂吧，就是k! = 1×2×⋯×(k−1)×k；

再往后简单推个两三个你就能知道答案啦，也就是输入x；只要输出x-1即可；

接下来代码实现部分就十分简单了，但也给了吧这题主要是理解部分难点

AC代码

#include<bits/stdc++.h>
using namespace std;
long long t,s;
int main()
{
    cin>>t;
	while(t--)
    {
        cin>>s;
        cout<<s-1<<endl;
    }
    return 0;
}