广搜+打表 POJ 1426 Find The Multiple
2016-06-11 19:23
169 查看
POJ 1426 Find The Multiple
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
Sample Output
Time Limit: 1000MS | Memory Limit: 10000K | |||
Total Submissions: 25734 | Accepted: 10613 | Special Judge |
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
/*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*/ /*---------------------*/ #define N 200 #include<iostream> using namespace std; #include<cstdio> #include<queue> queue<long long>que; int main() { freopen("1.out","w",stdout); for(int i=1;i<=N;++i) { que.push(1); while(!que.empty()) { long long x=que.front(); que.pop(); if(x%i==0) { cout<<x<<","; break; } que.push(x*10); que.push(x*10+1); } while(!que.empty()) que.pop(); } return 0; } /*-----------------------------------------------------------*/ #include<iostream> using namespace std; 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}; int main() { int n; while(cin>>n) { if(n==0) break; cout<<ans <<endl; } return 0; }
相关文章推荐
- Pycharm注册方式
- 用bat批处理不自动关闭cmd窗口
- 【JZOJ 3432】服务器 斜率优化DP常见问题&详细解答
- informix数据库选型及系统规划
- 【BZOJ 3907】【JZOJ 3431】网格
- gcc/g++等编译器 编译原理: 预处理,编译,汇编,链接各步骤详解
- 收集优秀文章,开发技巧,技术干货 (不断更新)
- 【折腾日记GEN8_0】预算及汇总
- redis内存数据库
- 【MyBatis学习04】mapper代理方法开发dao
- 【Leetcode】Design Twitter
- 测试者之间的闲聊
- 一篇作为日常笔记的文章(不定时更新)
- [BZOJ 3907] [JZOJ 3431] 网格
- NestedScrollView实现
- 随机行走
- 为Java程序中添加播放MIDI音乐功能
- 在Ubuntu 14.04 上安装网易云音乐
- jvm 内存划分 栈区 堆区 方法区
- xml用于非结构数据检索技术