Find The Multiple
| Time Limit: 1000MS | Memory Limit: 10000K | |||
| Total Submissions: 27988 | Accepted: 11621 | 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
挺难想到用bfs的，长度最长为100的数很容易往大数方向想。
想到用bfs对当前序列的后面添加0或1，直到当前序列取模为0。
再想到用同余模定理后就可以转化为将当前序列的模入队，之后只要在出队的模后添0或1后再取模入队，直到模为0。
不过新的问题在于如何把大数保存并输出。
可以发现事实上这道题的bfs是在对一棵节点只有0或1分支的二叉树作层次遍历。
所以我们可以根据二叉树的性质用存进队列的所有模（用一个数组保存）来输出大数的每一位。
具体看代码。
然后这道题如果用STL的queue，用C++编译是会TLE的，自己用数组实现的队列则不会，或者用G++编译也不会。
不过用STL的queue速度会慢上不少，我猜测可能部分时间用去分配空间了，毕竟自己用数组实现的话，数组也要开到百万以上。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#define FOP freopen("data.txt","r",stdin)
#define inf 0x3f3f3f3f
#define maxn 1000010
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;
int m;
int mods[maxn];
int ans[110];
int cot;
void bfs()
{
queue<int> Q;
Q.push(1);
while(!Q.empty())
{
int t = Q.front();
mods[cot++] = t;
Q.pop();
if(t == 0) break;
int t1 = (t * 10 + 0) % m;
int t2 = (t * 10 + 1) % m;
Q.push(t1);
Q.push(t2);
}
}
int main()
{
while(scanf("%d", &m) && m)
{
cot = 1;
int n = 0;
bfs();
cot--;
while(cot)
{
ans[n++] = cot % 2;
cot /= 2;
}
for(int i = n - 1; i >= 0; i--) printf("%d", ans[i]);
printf("\n");
}
return 0;
}