计蒜客 30990 - An Olympian Math Problem - [简单数学题][2018ICPC南京网络预赛A题]

Posted 2021-01-03 dilthey

题目链接：https://nanti.jisuanke.com/t/30990

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 * 3 * … * (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.

 

题意：

假设 $Sleft( n ight) = 1 imes 1! + 2 imes 2! + cdots + left( {n - 1} ight) imes left( {n - 1} ight)!$，求 $Sleft( n ight)$ 模 $n$ 的余数。

 

题解：

$egin{array}{l} 1 + Sleft( n ight) \ = 1 + 1 imes 1! + 2 imes 2! + cdots + left( {n - 1} ight) imes left( {n - 1} ight)! = 2 imes 1! + 2 imes 2! + cdots + left( {n - 1} ight) imes left( {n - 1} ight)! \ = 2! + 2 imes 2! + cdots + left( {n - 1} ight) imes left( {n - 1} ight)! = 3 imes 2! + cdots + left( {n - 1} ight) imes left( {n - 1} ight)! \ = 3! + 3 imes 3! + cdots + left( {n - 1} ight) imes left( {n - 1} ight)! = 4 imes 3! + cdots + left( {n - 1} ight) imes left( {n - 1} ight)! \ = cdots = left( {n - 1} ight)! + left( {n - 1} ight) imes left( {n - 1} ight)! = n imes left( {n - 1} ight)! = n! \ end{array}$

所以有 $Sleft( n ight)mod n = left( {n! - 1} ight)mod n = left( {n! + n - 1} ight)mod n = n!mod n + left( {n - 1} ight)mod n = n - 1$。

 

AC代码：

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