POJ 1426 Find The Multiple(bfs)
阿新 • • 發佈:2018-12-17
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
題意:給出一個整數n,(1 <= n <= 200)。求出任意一個它的倍數m,要求m必須只由十進位制的’0’或’1’組成。m第一個數字必須是1,且n不大於200,m不超過100個十進位制數字。
題解:搜尋,佇列都可以的,但是一定不要忘記將佇列清空。
#include<iostream>//佇列程式碼 #include<queue> using namespace std; #define ll long long ll n; queue<ll>q; void bfs() { q.push(1);//從1開始 while(!q.empty()){ ll front=q.front(); if(front%n==0) { cout<<front<<endl; return ; } q.pop(); for(int i=0;i<2;i++) { ll x=front*10+i; if(x%n==0) { cout<<x<<endl; return ; } q.push(x); } } } int main() { while(cin>>n&&n) { while(!q.empty()){ q.pop();} bfs(); } return 0; }
#include <iostream>//搜尋
#include <algorithm>
#include <cstdio>
using namespace std;
int n,m;
queue <long long> q;
void bfs()
{
q.push(1);
while(!q.empty())
{
long long t=q.front();
q.pop();
if(t%n==0)
{
cout<<t<<endl;
return ;
}
q.push(t*10);
q.push(t*10+1);
}
return ;
}
int main()
{
while(cin>>n&&n)
{
while(!q.empty())
{
q.pop();
}
bfs();
}
return 0;
}