poj 1426(同余搜索)

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 26926		Accepted: 11174		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

题意:找到一个由01组成的串m使得 m%n == 0
题解:同余定理.很强的搜索剪枝技巧。
同余剪枝,如果一个小数的余数和大数的余数相同,假设小点的数 a ,大点的为 b ,a%n==b%n => (a+c)%n == (b+c)%n ,我们假设 a+c是符合条件的,那么,b+c 也是符合条件的,所以我们可以直接不要
 b+c 了,这就是一个很巧妙并且很强的剪枝. 

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <stdlib.h>
#include <string>
#include <iostream>
using namespace std;
int n;
bool mod[]; ///这里是最强的剪枝,同余剪枝,如果一个小数的余数和大数的余数相同
               ///假设小点的数 a ,大点的为 b ,a%n==b%n => (a+c)%n == (b+c)%n ,我们假设 a+c是符合条件的,那么
               /// b+c 也是符合条件的,所以我们可以直接不要 b+c 了,这就是一个很巧妙并且很强的剪枝.
struct Node{
    int mod;
    string ans;
};
string bfs(){
    memset(mod,false,sizeof(mod));
    queue<Node> q;
    Node s;
    s.ans = "";
    s.mod = %n;
    q.push(s);
    while(!q.empty()){
        Node now = q.front();
        q.pop();
        if(now.mod==){
            return now.ans;
        }
        Node next;
        next.mod = (now.mod*+)%n;
        next.ans = now.ans+"";
        if(!mod[next.mod]){
            mod[next.mod] = true;
            q.push(next);
        }
        next.mod = (now.mod*)%n;
        next.ans = now.ans+"";
        if(!mod[next.mod]){
            mod[next.mod] = true;
            q.push(next);
        }
    }
}
int main(){

    while(scanf("%d",&n)!=EOF,n){
        cout<<bfs()<<endl;
    }
}