Codeforces Round #444 (Div. 2) B. Cubes for Masha 暴力枚举

Posted 2022-12-08 ProLightsfxjh

B. Cubes for Masha time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Absent-minded Masha got set of n cubes for her birthday.

At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural x such she can make using her new cubes all integers from 1 to x.

To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number.

The number can't contain leading zeros. It's not required to use all cubes to build a number.

Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations.

Input

In first line integer n is given (1 ≤ n ≤ 3) — the number of cubes, Masha got for her birthday.

Each of next n lines contains 6 integers aij (0 ≤ aij ≤ 9) — number on j-th face of i-th cube.

Output

Print single integer — maximum number x such Masha can make any integers from 1 to x using her cubes or 0 if Masha can't make even 1.

Examples input

3
0 1 2 3 4 5
6 7 8 9 0 1
2 3 4 5 6 7

output

87

input

3
0 1 3 5 6 8
1 2 4 5 7 8
2 3 4 6 7 9

output

98

Note

In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8.

Source

Codeforces Round #444 (Div. 2)

My Solution

题意：有n（1<= n <= 3）个骰子（每面标着数字0~9），要求找出最大的数x，满足1~x之间所有的数都可以用这最多n个骰子的正面表示出来。不能旋转，即不能用9表示6，反正亦然，且不要求所有的骰子都用上）。

暴力、枚举

首先用一个标记数组f，先全标记为false，之后把出现过的数都标记为true。

当n == 1的时候，只有枚举每一面，并标记即可。

当n == 2的时候，要枚举只用到1个骰子的情况和要用到2个骰子的情况，同时手动枚举全排列。

当n == 3的时候，要枚举只用到1个或者2个骰子的情况和要用到3个骰子的情况，同时手动枚举全排列。

详情请见代码。

时间复杂度 O（1000）

空间复杂度 O（1000）

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const int MAXN = 1e3 + 8;

bool f[MAXN];
int v[6][6];

int main()

    #ifdef LOCAL
    freopen("b.txt", "r", stdin);
    //freopen("b.out", "w", stdout);
    int T = 4;
    while(T--)
    #endif // LOCAL
    ios::sync_with_stdio(false); cin.tie(0);

    int n, i, j, k, ans = 0;
    cin >> n;
    for(i = 0; i < n; i++)
        for(j = 0; j < 6; j++)
            cin >> v[i][j];
        
    
    if(n == 1)
        for(j = 0; j < 6; j++)
            f[v[0][j]] = true;
        
    
    else if(n == 2)
        for(j = 0; j < 6; j++)
            f[v[0][j]] = true;
        
        for(j = 0; j < 6; j++)
            f[v[1][j]] = true;
        
        for(i = 0; i < 6; i++)
            for(j = 0; j < 6; j++)
                f[v[0][i]*10 + v[1][j]] = true;
                f[v[1][i]*10 + v[0][j]] = true;
            
        

    
    else
        for(j = 0; j < 6; j++)
            f[v[0][j]] = true;
        
        for(j = 0; j < 6; j++)
            f[v[1][j]] = true;
        
        for(j = 0; j < 6; j++)
            f[v[2][j]] = true;
        
        for(i = 0; i < 6; i++)
            for(j = 0; j < 6; j++)
                for(k = 0; k < 6; k++)
                    f[v[0][i]*100 + v[1][j]*10 + v[2][k]] = true;
                    f[v[1][i]*100 + v[0][j]*10 + v[2][k]] = true;

                    f[v[0][i]*100 + v[2][j]*10 + v[1][k]] = true;
                    f[v[2][i]*100 + v[0][j]*10 + v[1][k]] = true;

                    f[v[2][i]*100 + v[1][j]*10 + v[0][k]] = true;
                    f[v[1][i]*100 + v[2][j]*10 + v[0][k]] = true;
                
            
        
        for(i = 0; i < 6; i++)
            for(j = 0; j < 6; j++)
                f[v[0][i]*10 + v[1][j]] = true;
                f[v[1][i]*10 + v[0][j]] = true;

                f[v[0][i]*10 + v[2][j]] = true;
                f[v[2][i]*10 + v[0][j]] = true;

                f[v[2][i]*10 + v[1][j]] = true;
                f[v[1][i]*10 + v[2][j]] = true;
            
        

    
    for(i = 1; i < MAXN; i++)
        if(!f[i])
            break;
        
        ans = i;
    
    cout << ans << endl;


    #ifdef LOCAL
    memset(f, false, sizeof f);
    cout << endl;
    
    #endif // LOCAL
    return 0;

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