ACM-ICPC 2018 Nanjng network preliminaries (konjac brush title)

.A. An Olympian Math Problem

 

Alice, a student of grade 666, 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!:

k!=1×2×⋯×(k−1)×kk! = 1 \times 2 \times \cdots \times (k - 1) \times kk!=1×2××(k1)×k

We denote SSS:

S=1×1!+2×2!+⋯+S = 1 \times 1! + 2 \times 2! + \cdots +S=1×1!+2×2!++
(n−1)×(n−1)! (n - 1) \times (n-1)!(n1)×(n1)!

Then SSS module nnn is ____________

You are given an integer nnn.

You have to calculate SSS modulo nnn.

Input

The first line contains an integer T(T≤1000)T(T \le 1000)T(T1000), denoting the number of test cases.

For each test case, there is a line which has an integer nnn.

It is guaranteed that 2≤n≤10182 \le n\le 10^{18}2n1018.

Output

For each test case, print an integer SSS modulo nnn.

Hint

The first test is: S=1×1!=1S = 1\times 1!= 1S=1×1!=1, and 111 modulo 222 is 111.

The second test is: S=1×1!+2×2!=5S = 1\times 1!+2 \times 2!= 5S=1×1!+2×2!=5 , and 555 modulo 333 is 222.

 

Sample input

 

2
2
3

 

Sample Output

 

1
2 

Solution
play table x-1 was found to
push for a while do not know why
CODE:
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define int long long
 4 
 5 int cal(int x,int p)
 6 {
 7     int ans = 1;
 8     for(int i=1;i<=x;i++)
 9     ans*=i,ans%=p;
10     return ans;
11 }
12 void work(int x)
13 {
14     int ans = 0;
15     for(int i=1;i<x;i++)
16     ans =(ans+i*cal(i,x))%x;
17     cout<<x<<" "<<ans<<endl;
18 }
19 signed main()
20 {
21   int T;
22   for(cin>>T;T;T--)
23   {
24     int x;cin>>x;
25     cout<<x-1<<endl;
26   }
27   return 0;
28     for(int i=1;i<=100;i++)
29     work(i);
30 }
 

 


 

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Origin www.cnblogs.com/zhangbuang/p/11093542.html