题目链接：http://acm.hnu.cn/online/?action=problem&type=show&id=13341&courseid=0
| Joke with permutation |
| Time Limit: 3000ms, Special Time Limit:7500ms, Memory Limit:65536KB |
| Total submit users: 85, Accepted users: 57 |
| Problem 13341 : Special judge |
| Problem description |
|
Joey had saved a permutation of integers from 1 to n in a text file. All the numbers were written as decimal numbers without leading spaces. |
| Input |
|
The input file contains a single line with a single string — the Joey’s
permutation without spaces. |
| Output |
|
Write a line to the output file with the restored permutation. Don’t forget
the spaces! |
| Sample Input |
4111109876532 |
| Sample Output |
4 1 11 10 9 8 7 6 5 3 2 |
| Problem Source |
| NEERC 2014 |
| Submit Discuss Judge Status Problems Ranklist |
题目大意：将一串完整的字符串分成1~n个数。将空格补进去，并将其输出。
解题思路：1、注意这个n是可以求出来的
2、一个数字一个数字比较，只要满足这个数字小于n，还有保证这个数字没有访问过就ok啦
详见代码。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 5 using namespace std; 6 7 int n,flag; 8 char ch[110]; 9 int ans[110]; 10 bool vis[110]; 11 12 bool dfs(int i,int k) 13 { 14 if (flag==1) 15 return true; 16 if (k==n+1) 17 { 18 for (int i=1;i<k-1;i++) 19 { 20 printf ("%d ",ans[i]); 21 } 22 printf ("%d ",ans[k-1]); 23 flag=1; 24 return true; 25 } 26 if (ch[i]-'0'<=n&&!vis[ch[i]-'0']&&ch[i]-'0'>0) 27 { 28 ans[k]=ch[i]-'0'; 29 vis[ch[i]-'0']=1; 30 if(dfs(i+1,k+1)) return true; 31 vis[ch[i]-'0']=0; 32 } 33 if ((ch[i]-'0')*10+ch[i+1]-'0'<=n&&(ch[i]-'0')*10+ch[i+1]-'0'>9&&!vis[(ch[i]-'0')*10+ch[i+1]-'0']) 34 { 35 ans[k]=(ch[i]-'0')*10+ch[i+1]-'0'; 36 vis[(ch[i]-'0')*10+ch[i+1]-'0']=1; 37 if(dfs(i+2,k+1)) return true; 38 vis[(ch[i]-'0')*10+ch[i+1]-'0']=0; 39 } 40 return false; 41 } 42 43 int main() 44 { 45 while (scanf("%s",ch)!=EOF) 46 { 47 flag=0; 48 memset(ans,0,sizeof(ans)); 49 memset(vis,0,sizeof(vis)); 50 int len=strlen(ch); 51 if (len>9) 52 n=(len-9)/2+9; 53 else 54 n=len; 55 dfs(0,1);//dfs(); 56 } 57 return 0; 58 }