BFS广度优先搜索(9)(亦可以用DFS)--poj1426

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Find The Multiple

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu

Description

Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.

Input

The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.

Output

For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.

Sample Input

Sample Output

10100100100100100100111111111111111111

这道题题目大意是给出一个数n，找出一个数要求是n的倍数，并且这个数的十进制只由1和0组成，明显这样的数不止一个（如果，满足条件一定会有m×10也满足，故不止一种），题目要求输出任意一个满足该条件的m,对于给出的一个数2，可知2×5=10，故答案可以得出是10（当然，100，1000...也满足，但是special judge，只用输出一个满足条件的解），其他数据也同理。

思路很简单：每一位枚举0或1，位数逐渐往后增加即可，Bfs与Dfs都行。

Bfs代码：

#include<stdio.h>#include<queue>#define LL long long using namespace std;void Bfs(int n){queue<LL>q;q.push(1);     //从1开始枚举while(1){LL s=q.front();q.pop();if(s%n==0){printf("%lld\n",s);return;}q.push(s*10);    //位数增加q.push(s*10+1);}}int main(){int n;while(scanf("%d",&n)&&n){Bfs(n);}return 0;}

Dfs代码(位数大概搜到十九位左右，因为到20位时64位整数存不下)：

#include<iostream>#include<cstdlib>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std;bool found;void DFS(unsigned __int64 t ,int n,int k){if(found)return ;//如果已经发现了答案就没搜的必要了if(t%n==0){//发现答案，输出，标记变量该trueprintf("%I64u\n",t);found=true;return ;}if(k==19)//到第19层，回溯return ;DFS(t*10,n,k+1);    //搜索×10DFS(t*10+1,n,k+1);    //搜索×10+1}int main(){int n;while(cin>>n,n){found=false;//标记变量，当为true代表搜到了题意第一的mDFS(1,n,0);    //从1开始搜n的倍数，第三个参数代表搜的层数，当到第19层时返回（因为第20层64位整数存不下）}return 0;}

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