B. Cubes for Masha
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.
InputInput
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.
样例1
输入
3
0 1 2 3 4 5
6 7 8 9 0 1
2 3 4 5 6 7
输出
87
样例2
输入
3
0 1 3 5 6 8
1 2 4 5 7 8
2 3 4 6 7 9
输出
98
就是有n个骰子，每个面上有一个0-9的数字，求组起来的最长的连续数字，可以不用到全部骰子，例如样例1只有一个8，所以无法组合出88，而1-87都可以组合出，所有答案为87，这是一道水题，n范围是1-3，直接直接暴力就行，把所有组合存到一个数组里，再遍历找出无法组合出来的数。
#include
#include
#include
using namespace std;
int arr[5][10]；
int vis[100];
int main(){
int n;
scanf("%d",&n);
for(int i=0;i<n;i++){
for(int j=0;j<6;j++)
scanf("%d",&arr[i][j]);
}
for(int i=0;i<n;i++){
for(int j=0;j<6;j++){
vis[arr[i][j]]=1;
for(int l=0;l<n;l++){
if(l==i) continue;
for(int r=0;r<6;r++){
int x=arr[i][j]*10+arr[l][r];
vis[x]=1;
}
}
}
}
for(int i=1;i<100;i++)
if(!vis[i]){
printf("%d\n",i-1);
break;
}
return 0;}