poj_1426 Find The Multiple（bfs + 同余模定理）

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal 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

26190

Sample Output

10100100100100100100111111111111111111

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 longusing 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;}