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.
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.
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.
3
0 1 2 3 4 5
6 7 8 9 0 1
2 3 4 5 6 7
87
3
0 1 3 5 6 8
1 2 4 5 7 8
2 3 4 6 7 9
98
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.
思路：
暴力模拟，一个数字里面每个骰子有且至多能用一次
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 int weishu(int num) { 6 if(num>=1&&num<=9) return 1; 7 if(num>=10&&num<=99) return 2; 8 if(num>=100&&num<=999) return 3; 9 return 4; 10 } 11 bool exist(int ans[][10], int rec[], int len) { 12 if(len==1) { 13 int a=rec[1]; 14 if(ans[1][a]||ans[2][a]||ans[3][a]) return true; 15 } else if(len==2) { 16 int a=rec[1],b=rec[2]; 17 if((ans[1][a]&&ans[2][b]) 18 ||(ans[1][a]&&ans[3][b]) 19 ||(ans[2][a]&&ans[3][b]) 20 ||(ans[3][a]&&ans[2][b]) 21 ||(ans[3][a]&&ans[1][b]) 22 ||(ans[2][a]&&ans[1][b])) 23 return true; 24 } else if(len==3) { 25 int a=rec[1],b=rec[2],c=rec[3]; 26 if((ans[1][a]&&ans[2][b]&&ans[3][c]) 27 ||(ans[1][a]&&ans[3][b]&&ans[2][c]) 28 ||(ans[2][a]&&ans[3][b]&&ans[1][c]) 29 ||(ans[2][a]&&ans[1][b]&&ans[3][c]) 30 ||(ans[3][a]&&ans[1][b]&&ans[2][c]) 31 ||(ans[3][a]&&ans[2][b]&&ans[1][c])) 32 return true; 33 } 34 return false; 35 } 36 int main(){ 37 int ans[4][10],n,num,bns[4][10],rec[4]; 38 memset(ans,0,sizeof(ans)); 39 /*read*/ 40 scanf("%d",&n); 41 for(int i=1;i<=n;++i) { 42 int flag[10]={0}; 43 for(int j=0;j<6;++j) { 44 scanf("%d",&num); 45 if(flag[num]==0) { 46 ans[i][num]++; 47 flag[num]=1; 48 } 49 } 50 } 51 /*slove*/ 52 int result=0; 53 for(int i=1;i<=1000;++i) { 54 if(i==1000) { 55 result=999; 56 break; 57 } 58 int len=weishu(i); 59 int temp=i; 60 memcpy(bns,ans,sizeof(ans)); 61 memset(rec,-1,sizeof(rec)); 62 for(int j=1;j<=len;++j) { 63 int r=i%10; 64 rec[j]=r; 65 i=i/10; 66 } 67 i=temp; 68 if(!exist(ans,rec,len)) { 69 result=i-1; 70 break; 71 } 72 } 73 printf("%d ",result); 74 return 0; 75 }
优化暴力模拟，不可能出现比99大的数字，因为如果要出现比99大的数字，那么需要有1 1，2 2，3 3，4 4，5 5，6 6，7 7，8 8，9 9这样的话已经有18个数字，即使三个骰子也只是刚好18个数字。其中必不可少的是数字0，加上就是19。所以不可能出现比99大的数字。
1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() { 4 int n,ans[4][7],vis[100]; 5 memset(vis,0,sizeof(vis)); 6 scanf("%d",&n); 7 for(int i=1;i<=n;++i) { 8 for(int j=1;j<=6;++j) { 9 scanf("%d",&ans[i][j]); 10 } 11 } 12 for(int i=1;i<=n;++i) { 13 for(int j=1;j<=6;++j) { 14 vis[ans[i][j]]=1; 15 for(int k=1;k<=n;++k) { 16 if(k==i) continue; 17 for(int l=1;l<=6;++l) { 18 vis[ans[i][j]*10+ans[k][l]]=1; 19 } 20 } 21 } 22 } 23 for(int i=1;i<=99;++i) { 24 if(!vis[i]) { 25 printf("%d ",i-1); 26 break; 27 } 28 } 29 return 0; 30 }