POJ

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 32497 Accepted: 13595 Special Judge

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

26190

Sample Output

10100100100100100100111111111111111111

题意：

        输入一个数n，求另一个数是n的倍数并且只包含0,1

题解：

        这题略坑。看到他给的数字长度是100位，我直接就用了字符串存，bfs搜索（当时内心默认的是输出最小的满足条件的值），然而超时。。。后来用这个程序打表发现数据正好在long long范围内，然后直接dfs。。不需要求出最小的值，只要满足条件，值多么离谱都能过，所以搜索可以写的非常暴力。

TLE代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>#include<iostream>#include<queue>using namespace std;int n,flag;int check(string c){    int mod=0,len=c.length();    for(int i=0;i<len;i++)    {        mod=(mod*10+c[i]-48)%n;    }    if(mod==0)    {        cout<< c << endl;        return 1;    }    return 0;}void bfs(){    queue<string> q;    string s="";    q.push(s);    while(!q.empty())    {        string top=q.front(); q.pop();        q.push(top+"1");        if(check(top+"1"))        {            return;        }        if(top.compare("")==0)            continue;        q.push(top+"0");        if(check(top+"0"))        {            return;        }    }}int main(){    while(scanf("%d",&n) && n)    {        flag=0;        bfs();    }    return 0;}

AC代码：

#include<stdio.h>#include<string.h>int n,flag;void dfs(long long s,int step){    if(flag || step==19)        return;    if(s%n==0)    {        flag=1;        printf("%lld\n",s);        return;    }    dfs(s*10,step+1);    dfs(s*10+1,step+1);}int main(){    while(scanf("%d",&n) && n)    {        flag=0;        dfs(1,0);    }    return 0;}