Find The Multiple

0ms,飘过;这道题一开始用DFS做的TLE;想想2^200肯定超啊;后来像题中说枚举一个即可,说明有多个,于是缩小,一直网下缩,所成了WA,完了,得换路子,用BFS居然说超内存(MLE);
可能是分的情况太多而且存在大量的重复数据,      算,就200个数,打出来;然后过了!;


Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21769 Accepted: 8947 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

Source

Dhaka 2002
BFS(MLE)的代码;

#include <iostream>#include<queue>#include<cstdio>using namespace std;struct node{    int c[30];    int cnt;    int yu;};queue<node>Q;void BFS(int n){    int g;    node now ,tmp;    now.c[0]=1;    now.cnt=0;    now.yu=0;    while(!Q.empty())        Q.pop();    Q.push(now);    while(!Q.empty())    {        now=Q.front();        Q.pop();        g=(now.c[now.cnt]+now.yu*10)%n;        if(g==0)        {            printf("\"");            for(int i=0; i<=now.cnt; i++)            {                printf("%d",now.c[i]);            }            printf("\",");            return;        }        else        {            now.yu=g;        }        if(now.cnt>40)            continue;        for(int i=0; i<2; i++)        {            tmp=now;            if(i==0)            {                tmp.cnt++;                tmp.c[tmp.cnt]=0;                Q.push(tmp);            }            else            {                tmp.cnt++;                tmp.c[tmp.cnt]=1;                Q.push(tmp);            }        }    }}int main(){     freopen("out.txt","w",stdout);        for(int i=1;i<=202;i++)        BFS(i);    return 0;}

有上述打表AC的代码;

#include <iostream>#include<cstring>using namespace std; char  s[220][30]={"0","1","10","111","100","10","1110","1001","1000","111111111","10","11","11100","1001","10010","1110","10000","11101","1111111110","11001","100","10101","110","110101","111000","100","10010","1101111111","100100","1101101","1110","111011","100000","111111","111010","10010","11111111100","111","110010","10101","1000","11111","101010","1101101","1100","1111111110","1101010","10011","1110000","1100001","100","100011","100100","100011","11011111110","110","1001000","11001","11011010","11011111","11100","100101","1110110","1111011111","1000000","10010","1111110","1101011","1110100","10000101","10010","10011","111111111000","10001","1110","11100","1100100","1001","101010","10010011","10000","1111111101","111110","101011","1010100","111010","11011010","11010111","11000","11010101","1111111110","1001","11010100","10000011","100110","110010","11100000","11100001","11000010","111111111111111111","100","101","1000110","11100001","1001000","101010","1000110","100010011","110111111100","1001010111","110","111","10010000","1011011","110010","1101010","110110100","10101111111","110111110","100111011","111000","11011","1001010","10001100111","11101100","1000","11110111110","11010011","10000000","100100001","10010","101001","11111100","11101111","11010110","11011111110","11101000","10001","100001010","110110101","100100","10011","100110","1001","1111111110000","11011010","100010","1100001","11100","110111","11100","1110001","11001000","10111110111","10010","1110110","1010100","10101101011","100100110","100011","100000","11101111","11111111010","1010111","1111100","1111110","1010110","11111011","10101000","10111101","111010","1111011111","110110100","1011001101","110101110","100100","110000","100101111","110101010","11010111","11111111100","1001111","10010","100101","110101000","1110","100000110","1001011","1001100","1010111010111","110010","11101111","111000000","11001","111000010","101010","110000100","1101000101","1111111111111111110","111000011","1000","10010001","1010",};int main(){    int n;    while(cin>>n,n)    {        cout<<s[n]<<endl;    }    return 0;}