[Hangdian 100 questions] recursive solution special exercises (For Beginner) 2049 exam groom _ part of the wrong arrangement problem

Problem Description

First, dress each bride almost exactly, and sit in a row with large red hijabs randomly;
Then, let the bridegrooms find their brides. Each person is allowed to find only one, and no one is allowed to find one.
Finally, lift the hijab, if you find the wrong object, you must kneel in public ...

It seems that being a groom is not easy ...

Suppose there are N newly-married couples, among them M grooms find the wrong bride, how many possibilities are there for this to happen.

Input

The first line of the input data is an integer C, indicating the number of test instances, and then the C line of data, each line contains two integers N and M (1 <M <= N <= 20).

Output

For each test instance, please output the total number of possible occurrences of this situation, and the output of each instance occupies one line.

#include <iostream>

using namespace std;

int main()
{
    int c,i,j;
    int m,n;
    long long sm [21]; // The result of misalignment of m elements
    long long a[21];
    sm[0]=1;
    sm[1]=0;
    a[0]=1;
    a[1]=1;
    for(j=2;j<21;j++){
        sm[j]=(j-1)*(sm[j-1]+sm[j-2]);
        a[j]=j*a[j-1];
    }

    cin>>c;
    for(i=0;i<c;i++){
        cin>>n>>m;
        cout<<sm[m]*(a[n]/a[m]/a[n-m])<<endl;
    }
    return 0;
}

【postscript】

1. The reason is very simple. The combination number C (n, m) multiplied by m elements is staggered, but infinite WA is a bit painful. .

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