本文主要是介绍TOJ 3504 Repeatless Numbers / 深搜，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Repeatless Numbe
描述
Arepeatless number is a positive integer containing no repeated digits. For instance, the first 25 repeatless numbers are
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, …
Given an integer n, your goal is to compute thenth repeatless number.
输入
The input test file will contain multiple test cases, each consisting of a single line containing the integern, where 1 ≤n ≤ 1000000. The end-of-file is marked by a test case withn = 0 and should not be processed.
输出
For each input case, the program should print the nth repeatless number on a single line.
样例输入
25
10000
0
样例输出
27
26057
#include <stdio.h>
int a[1000010];
int t = 0;
int map[10];
void dfs(int sum,int k,int flag)//flag=0是表示0可以无限用 flag=1表示0只能用一次 00000001
{if(k == 8){a[t++] = sum;return;}if(t > 1000000)return;int i;for(i = 0; i < 10; i++){if(i == 0){if(!flag)//0000000就是这种情况 前面的0可以随便用 {dfs(0,k+1,0);}else//flag=1 你们说明 比如00000010最后这个0 只能用1次 {if(!map[i]){map[i]=1;dfs(sum*10+i,k+1,1);map[i]=0;}}}else{if(!map[i]){map[i]=1;dfs(sum*10+i,k+1,1);map[i]=0;}}}
}
int main()
{dfs(0,0,0);int n;while(scanf("%d",&n),n){printf("%d\n",a[n]);}return 0;
}
这篇关于TOJ 3504 Repeatless Numbers / 深搜的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
