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Find The Multiple (poj 1426 bfs)

2014-10-20 22:43 351 查看

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Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18454		Accepted: 7461		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

Sample Output

10
100100100100100100
111111111111111111

Source

Dhaka 2002

题意：输入整数n，求一个只由1和0两个数字组成的十进制整数m，m是整数n（200以内）的k倍，且要求k最小。

思路：运用同余定理+bfs,看这位大神这一题的同余定理。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

struct Node
{
int num;
int mod;
Node *pre;//记录路径
};

int N;
int visit[maxn];

void print(Node *x)//递归输出路径
{
if (x->pre==NULL)
printf("%d",x->num);
else
{
print(x->pre);
printf("%d",x->num);
}
}

void bfs()
{
int i;
queue<Node>Q;
memset(visit,0,sizeof(visit));
Node sst,*now=new Node;
visit[1]=1;
while (!Q.empty())
Q.pop();
sst.mod=1;sst.num=1;sst.pre=NULL;
Q.push(sst);
while (!Q.empty())
{
Node *st=new Node;
*st=Q.front();
Q.pop();
if (st->mod==0)
{
print(st);
printf("\n");
return ;
}
for (i=0;i<2;i++)
{
int xx=(st->mod*10+i)%N;
if (!visit[xx])
{
now->mod=xx;
now->num=i;
now->pre=st;
Q.push(*now);
visit[xx]=1;
}
}
}
return ;
}

int main()
{
while (scanf("%d",&N)&&N)
{
if (N==1)
{
printf("1\n");
continue;
}
bfs();
}
return 0;
}

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