【洛谷】P9752 [CSP-S 2023] 密码锁
【洛谷】P9752 [CSP-S 2023] 密码锁
题目传送门
题解
CCF数据稳定发挥
暴力出奇迹,这题可以说是这么多年 CSP-s 最水的 A 题了。
n ≤ 8 n \le 8 n≤8,这个数据很水,所以直接暴力枚举所有的锁的密码的情况,然后暴力比较是否能用 n n n 个密码转换而来。
代码
#include <bits/stdc++.h>
#define lowbit(x) x & (-x)
#define endl "\n"
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
namespace fastIO {inline int read() {register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f;}inline void write(int x) {if(x < 0) putchar('-'), x = -x;if(x > 9) write(x / 10);putchar(x % 10 + '0');return;}
}
using namespace fastIO;
int n, lockk[15][15][15][15][15], a, b, c, d, e, ans = 0;
int main() {//freopen(" .in", "r", stdin);//freopen(". out", "w", stdout);n = read();for(int i = 1; i <= n; i ++) {a = read(), b = read(), c = read(), d = read(), e = read();for(int j = 0; j <= 9; j ++) {lockk[j][b][c][d][e] ++;}for(int j = 0; j <= 9; j ++) {lockk[a][j][c][d][e] ++;}for(int j = 0; j <= 9; j ++) {lockk[a][b][j][d][e] ++;}for(int j = 0; j <= 9; j ++) {lockk[a][b][c][j][e] ++;}for(int j = 0; j <= 9; j ++) {lockk[a][b][c][d][j] ++;}for(int j = 1; j <= 9; j ++) {lockk[(a + j) % 10][(b + j) % 10][c][d][e] ++;}for(int j = 1; j <= 9; j ++) {lockk[a][(b + j) % 10][(c + j) % 10][d][e] ++;}for(int j = 1; j <= 9; j ++) {lockk[a][b][(c + j) % 10][(d + j) % 10][e] ++;}for(int j = 1; j <= 9; j ++) {lockk[a][b][c][(d + j) % 10][(e + j) % 10] ++;}}for(int A = 0; A <= 9; A ++) {for(int B = 0; B <= 9; B ++) {for(int C = 0; C <= 9; C ++) {for(int D = 0; D <= 9; D ++) {for(int E = 0; E <= 9; E ++) {if(lockk[A][B][C][D][E] == n) {ans ++;}}}}}}write(ans), putchar('\n');return 0;
}