广搜+打表 POJ 1426 Find The Multiple

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POJ 1426   Find The Multiple
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 25734   Accepted: 10613   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
 1 /*代码一:广搜加打表,速度极快,广搜每一位是0还是1,幸好题目的数据范围没有超过int*/
 2 /*---------------------*/
 3 #define N 200 
 4 #include<iostream>
 5 using namespace std;
 6 #include<cstdio>
 7 #include<queue>
 8 queue<long long>que; 
 9 int main()
10 {
11     freopen("1.out","w",stdout);
12     for(int i=1;i<=N;++i)
13     {
14         que.push(1);
15         while(!que.empty())
16         {
17             long long x=que.front();
18             que.pop();
19             if(x%i==0)
20             {
21                 cout<<x<<",";
22                 break;
23             }
24             que.push(x*10);
25             que.push(x*10+1);
26         }
27         while(!que.empty()) que.pop();
28     }
29     return 0;
30 }
31 /*-----------------------------------------------------------*/
32 #include<iostream>
33 using namespace std;
34 long long ans[201]={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};
35 int main()
36 {
37     int n;
38     while(cin>>n)
39     {
40         if(n==0) break;
41         cout<<ans[n]<<endl;
42     }
43     return 0;
44 }

 

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