（POJ1426）Find The Multiple <BFS 大数模>

Find The Multiple

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：

给你一个整数n,求他的一个十进制倍数m,要求m只由0，1组成，m最多100位；

分析：

由0,1组成的数从小到大：1，10，11，100，101，110，111.。。。

发现：1后面加个0 就变成了 10 ，后面加个1 就变成了11

10后面加个0 就变成了 100，后面加个1 就变成了 101

。。。。

所以我们可以按照这个规律很快的写出所有的有0，1组成的数。

但是：m可以是100位数。long long 只有64位。

我先按照long long 打了下表结果如下发现36，72等不行

long long 不行就说明m无法被直接表示出来，所以就只能用mod运算了，明确用字符数组来表示m。

然后按照变换规律进行BFS即可。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
int n;
bool v[210];

struct node
{
char m[110];//m值
int len;//长度
int mod;//取模的结果
};

void bfs()
{
queue<node> q;
while(!q.empty()) q.pop();
memset(v,0,sizeof(v));
node now;
now.len = 1;
now.m[0] = '1';
now.mod = 1 % n;
v[now.mod] = 1;
q.push(now);
while(!q.empty())
{
now = q.front(); q.pop();
if(now.mod == 0)
{
for(int i=0;i<now.len;i++)
printf("%c",now.m[i]);
printf("\n");
return ;
}
now.m[now.len] = '0';
now.len++;
now.mod = (now.mod * 10) % n;
if(!v[now.mod]) q.push(now);
v[now.mod] = 1;

now.m[now.len-1] = '1';
now.mod = (now.mod + 1) % n;
if(!v[now.mod]) q.push(now);
v[now.mod] = 1;
}
}

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(scanf("%d",&n)!=EOF && n)
{
bfs();
}
return 0;
}

long long 打表代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long LL;
int n;

void bfs()
{
queue<LL> q;
while(!q.empty()) q.pop();
LL now,ne;
now = 1;
q.push(now);
while(!q.empty())
{
now = q.front(); q.pop();
if(now % n == 0)
{
printf("%d\n",now);
return;
}
for(int i=0;i<2;i++)
{
ne = now * 10 + i;
q.push(ne);
}
}
}

int main()
{
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
while(scanf("%d",&n)!=EOF && n)
{
bfs();
}
return 0;
}