#include <bits/stdc++.h> int n, shuru; int a[210][1010]; int main(){ std::cin >> n; for(int i = 1; i <= n; i ++) { for(int j = 1; j <= n; j ++) { std::cin >> shuru; } } a[1][0] = 1; a[1][1] = 0; a[2][0] = 1; a[2][1] = 1; for(int i = 3; i <= n; i ++) { int jw = 0; for(int j = 1; j <= 1000; j ++) { a[i][j] = a[i - 1][j] + a[i - 2][j] + jw; jw = 0; jw = a[i][j] / 10; a[i][j] = a[i][j] % 10; } a[i][0] = a[i - 1][0]; for(int j = 1; j <= 1000; j ++) { a[i][j] = a[i][j] * (i - 1) + jw; jw = a[i][j] / 10; a[i][j] = a[i][j] % 10; } } int flag = 0; for(int i = 1000; i >= 1; i --) { if(flag == 1 || !(a[n][i] == 0)) { flag = 1; std::cout << a[n][i]; } } return 0; }

这是一道错排问题，不知道的可以先做一下P1985(直接复制过来能得60分)

递推公式：f (i)=(i - 1) * (f (i - 1) + f (i - 2))

代码如下：

#include <bits/stdc++.h>
int n, shuru;
int a[210][1010];
int main(){
    std::cin >> n;
    for(int i = 1; i <= n; i ++)
    {
    	for(int j = 1; j <= n; j ++)
    	{
    		std::cin >> shuru;
		}
	}
    a[1][0] = 1; a[1][1] = 0;
    a[2][0] = 1; a[2][1] = 1;
    for(int i = 3; i <= n; i ++)
    {
    	int jw = 0;
    	for(int j = 1; j <= 1000; j ++)
    	{
    		a[i][j] = a[i - 1][j] + a[i - 2][j] + jw;
    		jw = 0;
			jw = a[i][j] / 10;
    		a[i][j] = a[i][j] % 10;
		}
		a[i][0] = a[i - 1][0];
		for(int j = 1; j <= 1000; j ++)
		{
			a[i][j] = a[i][j] * (i - 1) + jw;
			jw = a[i][j] / 10;
			a[i][j] = a[i][j] % 10;
		}
	}
	int flag = 0;
	for(int i = 1000; i >= 1; i --)
	{
		if(flag == 1 || !(a[n][i] == 0))
		{
			flag = 1;
			std::cout << a[n][i];
		}
	}
    return 0;
}

时间复杂度O(2000*n)

#include <bits/stdc++.h>
int n, shuru;
int a[210][1010];
int main(){
    std::cin >> n;
    for(int i = 1; i <= n; i ++)
    {
    	for(int j = 1; j <= n; j ++)
    	{
    		std::cin >> shuru;
		}
	}
    a[1][0] = 1; a[1][1] = 0;
    a[2][0] = 1; a[2][1] = 1;
    for(int i = 3; i <= n; i ++)
    {
    	int jw = 0;
    	for(int j = 1; j <= 1000; j ++)
    	{
    		a[i][j] = a[i - 1][j] + a[i - 2][j] + jw;
    		jw = 0;
			jw = a[i][j] / 10;
    		a[i][j] = a[i][j] % 10;
		}
		a[i][0] = a[i - 1][0];
		for(int j = 1; j <= 1000; j ++)
		{
			a[i][j] = a[i][j] * (i - 1) + jw;
			jw = a[i][j] / 10;
			a[i][j] = a[i][j] % 10;
		}
	}
	int flag = 0;
	for(int i = 1000; i >= 1; i --)
	{
		if(flag == 1 || !(a[n][i] == 0))
		{
			flag = 1;
			std::cout << a[n][i];
		}
	}
    return 0;
}