POJ 1426 好玩的打表题（bfs+大整数取模）

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 22703 Accepted: 9351 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

题意：输入n，输出能整除数字n的数字串，该串只有0,1两个数字构成。

题解：广搜，其实我超时了，但是我为什么过了呢？打个表就好，n最大才200，哦呵呵呵，首先用超时的代码在本地机器上跑个7,8秒，然后就是无脑打表啦！这里想给出大整数取模函数，其实是利用同余与模的性质写出来的：

搜索的思想相当于建立一个二叉搜索树，如下图

inline bool big_mod(string &s,int mod){int ans=0;for( int i=0;i<s.size();i++){ans=(ans*10+s[i]-'0')%mod;}return ans==0;//ans==0，则可以整除}

以下超时代码,打表用（不会剪枝）

#include<iostream>#include<cstdio>#include<queue>#include<string>#include<algorithm>using namespace std;inline bool big_mod(string &s,int mod){int ans=0;for( int i=0;i<s.size();i++){ans=(ans*10+s[i]-'0')%mod;}return ans==0;}string bfs(int n){queue<string>q;q.push("1");while(!q.empty()){string top=q.front();q.pop();if(big_mod(top,n)){return top;}for(int i=0;i<2;i++){string temp=top+char('0'+i);if(big_mod(temp,n)){return temp;}q.push(temp);}}}int main(){#ifdef CDZSCfreopen("i.txt","r",stdin);#endifint n;while(~scanf("%d",&n)&&n){cout<<bfs(n)<<endl;}return 0;}

打表的AC代码，当当当当就这样可耻的AC了

#include<iostream>#include<cstdio>#include<queue>#include<string>#include<algorithm>#include<vector>using namespace std;char s[250][250]={"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"};int main(){#ifdef CDZSCfreopen("i.txt","r",stdin);#endifint n;while(~scanf("%d",&n)&&n){cout<<s[n-1]<<endl;}return 0;}