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.

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.

题解：

自己写写错了。注意题意。直接或运算暴力即可。还看了别人的。

代码：

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int weishu(int num){    if(num>=1&&num<=9) return 1;    else if(num>=10&&num<=99) return 2;    else if(num>=100&&num<=999) return 3;    return 4;}bool exist(int ans[][10], int rec[], int len) {    if(len==1) {        int a=rec[1];        if(ans[1][a]||ans[2][a]||ans[3][a]) return true;    } else if(len==2) {        int a=rec[1],b=rec[2];        if((ans[1][a]&&ans[2][b])            ||(ans[1][a]&&ans[3][b])            ||(ans[2][a]&&ans[3][b])            ||(ans[3][a]&&ans[2][b])            ||(ans[3][a]&&ans[1][b])            ||(ans[2][a]&&ans[1][b]))        return true;    } else if(len==3) {        int a=rec[1],b=rec[2],c=rec[3];        if((ans[1][a]&&ans[2][b]&&ans[3][c])            ||(ans[1][a]&&ans[3][b]&&ans[2][c])            ||(ans[2][a]&&ans[3][b]&&ans[1][c])            ||(ans[2][a]&&ans[1][b]&&ans[3][c])            ||(ans[3][a]&&ans[1][b]&&ans[2][c])            ||(ans[3][a]&&ans[2][b]&&ans[1][c]))        return true;    }    return false;}int main(){    int ans[4][10],n,num,rec[4];    memset(ans,0,sizeof(ans));    /*read*/    scanf("%d",&n);    for(int i=1;i<=n;++i)    {        int flag[10]={0};        for(int j=0;j<6;++j)        {            scanf("%d",&num);            if(flag[num]==0)            {                ans[i][num]++;                flag[num]=1;            }        }    }    /*slove*/    int result=0;    for(int i=1;i<=1000;++i)    {        if(i==1000) {            result=999;            break;        }        int len=weishu(i);        int temp=i;        memset(rec,-1,sizeof(rec));        for(int j=1;j<=len;++j) {            int r=i%10;            rec[j]=r;            i=i/10;        }        i=temp;        if(!exist(ans,rec,len)) {            result=i-1;            break;        }    }    printf("%d\n",result);    return 0;}