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[深度优先搜索] POJ 1426 Find The Multiple
Find The Multiple
Time Limit: 1000MS | Memory Limit: 10000K | |||
Total Submissions: 28550 | Accepted: 11828 | 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
Source
Dhaka 2002
原题大意:找出任意一个只由0与1组成的数,使得其为n的倍数。
解题思路:问题在于对问题的优化。100位数字太长了(当然,实际到不了100位,最高19位左右),若是找余数的状态,又很难数清有多少种。于是可以先来一个裸的搜索先判最长位数,然后用深搜或广搜解题。
#include<stdio.h> int n,ans[400]; bool found; void dfs(int mod,int dep) { if(found || dep>19) return; if(mod==0) { for(int i=0;i<=dep;++i) printf("%d",ans[i]); printf("\n"); found=true; return; } ans[dep+1]=1; dfs((mod*10+1)%n,dep+1); //mod的运算规律,可以这样想:假设X为当前位以前所有的数,那么当前数为X*10+i(i=0/1),对其取n的余数,(X%n*10%n+i%n)%n,递推即可。 ans[dep+1]=0; dfs((mod*10)%n,dep+1); } int main() { int i; while(~scanf("%d",&n)) { if(n==0) break; found=false; ans[0]=1; dfs(1%n,0); } return 0; }
[深度优先搜索] POJ 1426 Find The Multiple
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