POJ 1426 BFS

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 35790		Accepted: 14957		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

2
6
19
0

Sample Output

10
100100100100100100
111111111111111111

Source

Dhaka 2002

题意：给定一个数，找到由一个01串使得是它的倍数。

分析：直接bfs每次加0加1，同时每次取模，判断是不是为0就可以了。

代码：

////#include "bits/stdc++.h"
#include "cstdio"
#include "map"
#include "set"
#include "cmath"
#include "queue"
#include "vector"
#include "string"
#include "cstring"
#include "time.h"
#include "iostream"
#include "stdlib.h"
#include "algorithm"
#define db double
#define ll long long
#define vec vector<ll>
#define Mt  vector<vec>
#define ci(x) scanf("%d",&x)
#define cd(x) scanf("%lf",&x)
#define cl(x) scanf("%lld",&x)
#define pi(x) printf("%d
",x)
#define pd(x) printf("%f
",x)
#define pl(x) printf("%lld
",x)
#define rep(i, x, y) for(int i=x;i<=y;i++)
const int N   = 1e6 + 5;
const int mod = 1e9 + 7;
const int MOD = mod - 1;
const db  eps = 1e-18;
const db  PI  = acos(-1.0);
using namespace std;
int n;
struct P
{
    int s[20];
    int ans,cnt;
};
bool v[205];
//int R()
//{
//    int x=0,f=1;char ch=getchar();
//    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
//    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
//    return x*f;
//}
inline void bfs(int x)
{
    queue<P> q;
    P p;
    memset(p.s,0,sizeof(p.s));
    p.s[0]=1;
    p.cnt=1;
    p.ans=1;
    v[p.ans]=1;
    q.push(p);
    while(q.size())
    {
        P p1=q.front();
        q.pop();
        if(!p1.ans){
            for(int i=0;i<p1.cnt;i++)
                printf("%d",p1.s[i]);
            puts("");
            return;
        }
        for(int i=0;i<=1;i++){
            P p2;
            for(int ii=0;ii<p1.cnt;ii++) p2.s[ii]=p1.s[ii];
            p2.cnt=p1.cnt,p2.ans=p1.ans;
            p2.ans=(p2.ans*10+i)%x;
            if(v[p2.ans]==1) continue;//限制队列中元素的个数，否则会MLE
            v[p2.ans]=1;
            p2.s[p2.cnt++]=i;
            q.push(p2);
        }
    }
}
int main()
{
    while(scanf("%d",&n)==1&&n)
    {
        memset(v,0, sizeof(v));
        bfs(n);
    }
}