TOJ 3504 Repeatless Numbers / 深搜

Repeatless Numbe

时间限制(普通/Java):1000MS/3000MS     运行内存限制:65536KByte
 

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

25100000

样例输出

2726057

#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;}