Codeforces Round #444 (Div. 2) B题 （887B）

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 a_ij (0 ≤ a_ij ≤ 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.

 题意：从色子的数字中选数字（每个色子上的数字只能选一个），然后组成数字 ， 组成 连续的1---x  ， 问 x 为多少 。

 思路：直接暴力

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std ;

#define maxn 100
int num[10][maxn] ;
bool visit[maxn*100] ;

int main() {
    int n ;
    while(~scanf("%d" , &n)) {
        for(int i=1 ; i<=n ; i++) {
            for(int j=1 ; j<=6 ; j++) {
                scanf("%d" , &num[i][j]) ;
            }
        }
        memset(visit , false , sizeof(visit)) ;

        for(int i=1 ; i<=n ; i++) {
            for(int j=1 ; j<=6 ; j++) {
                visit[num[i][j]] = true ;
                for(int k=1 ; k<=n ; k++) {
                    if(k==i) continue ; 
                    for(int l = 1 ; l<=6 ; l++){
                        visit[num[i][j]*10+num[k][l]] = true ; 
                        for(int m = 1 ; m<=n ; m++){
                            if(m==i||m==k) continue ; 
                            for(int g = 1 ; g<=6 ; g++){
                                visit[num[i][j]*100+num[k][l]*10+num[m][g]] = true ; 
                            }
                        }
                    }
                }
            }
        }
        
        for(int i=1 ; i<=10*maxn ; i++){
            if(!visit[i]){
                printf("%d
" , i-1) ; 
                break ; 
            }
        }
    }

    return 0 ;
}

其实仔细想想并不需要那么多的循环 因为 最多三个色子 ， 总共 18 个数字，如果要组成100 则必须要能组成 11 22 33 44 55 66 77 88 99 ，这就已经达到 18 了 ，所以不会组成

三位数字

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std ;

#define maxn 150 
int num[10][10] ; 
bool visit[maxn ] ; 
int n ;

int main() {
    
    while(~scanf("%d" , &n)){
        for(int i=1 ; i<=n ; i++){
            for(int j=1 ; j<=6 ; j++){
                scanf("%d" , &num[i][j]) ; 
            }
        }
        
        for(int i=1 ; i<= n ;i++){
            for(int j=1 ; j<= 6 ; j++){
                visit[num[i][j]] = true ; 
                for(int k=1 ; k<= n ; k++){
                    if(k == i){
                        continue ; 
                    }
                    for(int l = 1 ; l<=6 ; l++){
                        visit[num[i][j]*10 + num[k][l]] = true ; 
                    }
                }
            }
        }
        
        for(int i=1 ; i<=maxn ; i++){
            if(!visit[i]){
                printf("%d
" , i-1) ; 
                break ; 
            }
        }
    }    

    return 0 ; 
}