题解 | #牛牛的排列数#

#include <stdio.h>

int A_(int n, int m) {
    if (n < m)
        return 0;
    else {
        long long a1 = 1;
        long long a2 = 1;
        int i = 0;
        for (i = n; i > 0; a1 *= i, i--);
        for (i = n - m; i > 0; a2 *= i, i--);
        return (a1 / a2);
    }
}

int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    printf("%d", A_(n, m));
    return 0;
}