poj-1426 Find The Multiple BFS打表

题目链接

poj-1426 Find The Multiple

题目

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 37783		Accepted: 15777		Special Judge

DescriptionGiven 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.InputThe 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.OutputFor 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 Input2
6
19
0Sample Output10
100100100100100100
111111111111111111SourceDhaka 2002

题意

翻译的比较绕，给一个200以内的正整数，要求在该数的所有倍数中找到一个数，但是每位只能由0或者1组成。

题解

比较奇葩的题，如果不是划分到搜索，根本想不到用搜索来写。。。简单地说，由于在十进制中只能出现1跟0，同时又要求是原数的倍数。所以在数据量特别大的时候要利用字符串，并且完成大数的取模运算。但是该题数据量不大（Discuss有用unsigned long long）用long long即可完美覆盖数据。可以使用bfs搜索所有可能，对单一位而言，下一位有0与1两种可能。所以每次搜索时2步选择即可（同时因为这个特性，完全可以利用dp）。但是实际上在不剪枝的情况下bfs超时，但是由于整体数据很小，只有200个，所以完全可以利用bfs得到200个数据，然后打表解决问题。
bfs部分：
由于答案必定是1与0组合得到的十进制数，所以可以将得到答案的过程看成搜索的过程。首位当然不能是0，所以队列的第一位当然是1，然后进入bfs函数，拿出头后，首先判断是否取余为0，是则搜索成功（special judge-找出一个即可）。下一位有2种情况——0或者1，那么就将这2种情况分别加入队列，就是：que.push(n*10);
que.push(n*10+1);这样，我们最终就能输出一个找到的数据。
由于bfs一定是超时的，所以我们选择打表，输出格式上进行选择，可以在本地上跑这个bfs的算法，生成表，然后将表存放在数组中，即可解决问题。

C++ 生成表数据代码

#include<iostream>
#include<vector>
#include<string>
#include<math.h>
#include<algorithm>
#include<map>
#include<utility>
#include<queue>
#include<memory>
#define TIME std::ios::sync_with_stdio(false)
#define LL long long
#define MAX 310
#define INF 0x3f3f3f3f

using namespace std;

long long bfs(queue<long long> que,int flag){
while(!que.empty()){
long long now = que.front();
que.pop();
if(now % flag == 0) return now;
que.push(now*10);
que.push(now*10+1);
}
return 0;
}

int main() {
TIME;
for(int i = 1;i < 201;i++){
queue<long long> que;
que.push(1);
long long ans = bfs(que,i);
cout << "\"" << ans << "\"" << ",";
}
cout << endl << "----end" << endl;

system("pause");
return 0;
}这样，我们将输出的数据全部复制，将最后一个","删除后就是表了。

C++ 打表代码 16ms

#include<iostream>
#include<vector>
#include<string>
#include<math.h>
#include<algorithm>
#include<map>
#include<utility>
#include<queue>
#include<memory>
#define TIME std::ios::sync_with_stdio(false)
#define LL long long
#define MAX 310
#define INF 0x3f3f3f3f

using namespace std;

string ans[200] = { "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" };

int main() {
int n;
while (cin >> n) {
if (n == 0) break;
cout << ans[n - 1] << endl;
}
system("pause");
return 0;
}对应着数组输出即可。