zoukankan      html  css  js  c++  java
  • Joseph(约瑟夫环)

    Joseph

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 2094    Accepted Submission(s): 1273


    Problem Description
    The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n, standing in circle every mth is going to be executed and only the life of the last remaining person will be saved. Joseph was smart enough to choose the position of the last remaining person, thus saving his life to give us the message about the incident. For example when n = 6 and m = 5 then the people will be executed in the order 5, 4, 6, 2, 3 and 1 will be saved.

    Suppose that there are k good guys and k bad guys. In the circle the first k are good guys and the last k bad guys. You have to determine such minimal m that all the bad guys will be executed before the first good guy.
     
    Input
    The input file consists of separate lines containing k. The last line in the input file contains 0. You can suppose that 0 < k < 14.
     
    Output
    The output file will consist of separate lines containing m corresponding to k in the input file.
     
    Sample Input
    3 4 0
     
    Sample Output
    5 30
    题解:F(i)=(F(i-1)+m-1)%(n-i+1)由于去的后k个所以当F小于k就不符合。需要打表,否则超时;
    代码:
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int k;
    int dp[15];
    bool solve(int m){
    	int cur=0;
    	for(int i=1;i<=k;i++){
    		if((cur+m-1)%(2*k-i+1)<k)
    			return false;
    		cur=(cur+m-1)%(2*k-i+1);
    	}
    	return true;
    }
    int main(){
    	for(k=1;k<15;k++){
    		int i;
    		for(i=1;;i++){
    			if(solve(i))break;
    		}
    		dp[k]=i;
    	}
    	while(~scanf("%d",&k),k){
    		printf("%d
    ",dp[k]);
    	}
    	return 0;
    }
    
  • 相关阅读:
    Codeforces Round #365 (Div. 2) D
    Codeforces Round #414 C. Naming Company
    Codeforces Round #365 (Div. 2) B
    LA 6893 The Big Painting(矩阵Hash)
    Gym100783C Golf Bot(FFT)
    POJ 2481 Cows(树状数组)
    POJ 2352 Stars
    POJ 2299 Ultra-QuickSort(树状数组+离散化)
    LightOJ 1341 Aladdin and the Flying Carpet(唯一分解定理)
    LightOJ 1356 Prime Independence(质因数分解+最大独立集+Hopcroft-Carp)
  • 原文地址:https://www.cnblogs.com/handsomecui/p/5125189.html
Copyright © 2011-2022 走看看