计蒜客 30990.An Olympian Math Problem-数学公式题 (ACM-ICPC 2018 南京赛区网络预赛 A)

 A. An Olympian Math Problem

• 54.28%
•  1000ms
•  65536K

 

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!:

k! = 1 imes 2 imes cdots imes (k - 1) imes kk!=1×2×⋯×(k−1)×k

We denote SS:

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

Then SS module nn is ____________

You are given an integer nn.

You have to calculate SS modulo nn.

Input

The first line contains an integer T(T le 1000)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 le nle 10^{18}2≤n≤1018.

Output

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

Hint

The first test is: S = 1 imes 1!= 1S=1×1!=1, and 11 modulo 22 is 11.

The second test is: S = 1 imes 1!+2 imes 2!= 5S=1×1!+2×2!=5 , and 55 modulo 33 is 22.

样例输入复制

2
2
3

样例输出复制

1
2

题目来源

ACM-ICPC 2018 南京赛区网络预赛

 

 

题意很好理解。

直接代码

代码:

//A-数学公式
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;

const double PI=acos(-1.0);
const double eps=1e-6;
const ll mod=1e9+7;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//公式为1*1！+2*2！+3*3!+...+n*n!=(n+1)!-1,本题为(n!-1)%n
//因为n!-1=(n-1)*n-1=(n-2)*n+n-1,所以[(n-2)*n+(n-1)]%n=n-1
//因n*n!=(n+1-1)n!=(n+1)n!-n!=（n+1)!-n!
//所以：1*1!=2!-1!
//2*2!=3!-2!
//3*3!=4!-3!
//.
//n*n!=(n+1)!-n!
//相加后有：1*1!+2*2!+3*3!+.+n*n!=（n+1)!-1
//1*1!+2*2!+3*3!+.+n*n!=（n+1)!-1
//把最后一项拆开来,变成(n+1-1)n!=(n+1)n!-n!

int main()
{
    int t;
    cin>>t;
    while(t--){
        ll n;
        cin>>n;
        cout<<n-1<<endl;
    }
}

溜了，一会贴一下线段树的题目。