[CSP-S 2023] 密码锁

题目链接：

CSP-S2023-T1

解题思路：

这题也太水了，数据甚至 $n<9$，而且一眼暴力，考场直接秒 $A$。

首先我们发现，在 $n=1$ 时，密码锁的可能的转动只有 $81$ 种，于是我们就可以骗分拿基础分：

if(n == 1){
    printf("81\n");
    return 0;
}

然后就可以直接大暴力：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
using namespace std;
const int MAXN = 10;
const int MAXM = 10000;
int n;
int a[MAXN][MAXN];
int w[MAXN][MAXM];
bool flag[100005][9];
int cnt[1000005];
int main(){
	// freopen("lock.in", "r", stdin);
	// freopen("lock.out", "w", stdout);
	scanf("%d", &n);
	for(int i = 1;i <= n;i++){
		for(int j = 1;j <= 5;j++){
			scanf("%d", &a[i][j]);
		}
	}
	if(n == 1){
		int num = 81;
		printf("%d\n", num);
		return 0;
	}
	for(int i = 1;i <= n;i++){
		int cc = 1;
		for(int j = 1;j <= 5;j++){
			for(int k = 1;k <= 9;k++){
				int num;
				if(k + a[i][j] > 9){
					num = k + a[i][j] - 10;
				}else{
					num = k + a[i][j];
				}
				if(j == 1){
					w[i][cc] = num * 10000 + a[i][2] * 1000 + a[i][3] * 100 + a[i][4] * 10 + a[i][5];
				}else if(j == 2){
					w[i][cc] = a[i][1] * 10000 + num * 1000 + a[i][3] * 100 + a[i][4] * 10 + a[i][5];
				}else if(j == 3){
					w[i][cc] = a[i][1] * 10000 + a[i][2] * 1000 + num * 100 + a[i][4] * 10 + a[i][5];
				}else if(j == 4){
					w[i][cc] = a[i][1] * 10000 + a[i][2] * 1000 + a[i][3] * 100 + num * 10 + a[i][5];
				}else if(j == 5){
					w[i][cc] = a[i][1] * 10000 + a[i][2] * 1000 + a[i][3] * 100 + a[i][4] * 10 + num;
				}
				cc++;
			}
		}
		for(int j = 1;j <= 4;j++){
			for(int k = 1;k <= 9;k++){
				int num1,num2;
				if(k + a[i][j] > 9){
					num1 = k + a[i][j] - 10;
				}else{
					num1 = k + a[i][j];
				}
				if(k + a[i][j + 1] > 9){
					num2 = k + a[i][j + 1] - 10;
				}else{
					num2 = k + a[i][j + 1];
				}
				
				if(j == 1){
					w[i][cc] = num1 * 10000 + num2 * 1000 + a[i][3] * 100 + a[i][4] * 10 + a[i][5];
				}
				else if(j == 2){
					w[i][cc] = a[i][1] * 10000 + num1 * 1000 + num2 * 100 + a[i][4] * 10 + a[i][5];
				}
				else if(j == 3){
					w[i][cc] = a[i][1] * 10000 + a[i][2] * 1000 + num1 * 100 + num2 * 10 + a[i][5];
				}
				else if(j == 4){
					w[i][cc] = a[i][1] * 10000 + a[i][2] * 1000 + a[i][3] * 100 + num1 * 10 + num2;
				}
				cc++;
			}
		}
	}
	
	for(int i = 1;i <= n;i++){
		for(int j = 1;j <= 81;j++){
			cnt[w[i][j]]++;
		}
	}
	int ans = 0;
	for(int i = 1;i <= 1;i++){
		for(int j = 1;j <= 81;j++){
			if(cnt[w[i][j]] == n){
				ans ++;
			}
		}
	}
	printf("%d\n", ans);
	return 0;				
}

$END$