题目链接：https://vjudge.net/problem/POJ-1426
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.
广搜水题。
题面翻译：
给定1个n，求一个数M满足：①M是n的倍数 ②M每个数位上只能是1或者0
分析：
这题看着很恐怖，因为题目说M不超过100位，但其实因为n的范围为1到200，我们在long long范围内就能找到对应的M；
用广搜是因为广搜能保证找到的数尽量小。
吐槽一下，用bfs不知道为啥会超时，调试的时候根本不会超时。然后用了打表过了。
打表代码：
#include <cstdio> #include <queue> #include <string.h> #include <iostream> using namespace std; unsigned long long excel[300]={ }; unsigned long long n; void bfs(){ queue<unsigned long long> q; q.push(1); while(!q.empty()){ unsigned long long temp=q.front(); q.pop(); if(temp%n==0){ printf("%llu,",temp); return; } unsigned long long temp1=temp*10; unsigned long long temp2=temp*10+1; q.push(temp1); q.push(temp2); } } int main() { //while(~scanf("%llu",&n)&&n) //bfs(); for(int i=1;i<=200;i++){ n=i; bfs(); } return 0; }打表结果存到excel数组就好了。
AC代码：
#include <cstdio> #include <queue> #include <string.h> #include <iostream> using namespace std; long long excel[300]={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}; int n; int main() { while(~scanf("%d",&n)&&n) printf("%lld\n",excel[n]); return 0; }
