hdu2574(一个数的不同的质数因子个数)

最新推荐文章于 2024-08-13 09:45:00 发布

原创最新推荐文章于 2024-08-13 09:45:00 发布 · 886 阅读

0 ·

CC 4.0 BY-SA版权

此算法描述了HduGirls'Day活动中一种特殊的投票策略：参与者会将票投给编号拥有最多不同质因数的女孩，若多个女孩质因数数量相同，则选择名字字典序最小者。

Hdu Girls' Day

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1045 Accepted Submission(s): 333

Problem Description

Hdu Girls' Day is a traditional activity in Hdu. Girls in Hdu participate in the activity and show their talent and skill. The girls who win in the activity will become the Hdu's vivid ambassadors(形象大使). There are many students in Hdu concern the activity. Now it's the finally competition to determine who will be the Hdu's vivid ambassadors. The students vote for the girl they prefer. The girl who has the most number of votes will be the first. You as a student representing Hdu Acm team has a chance to vote. Every girl who participates in the activity has an unique No. and name. Because you very like prime number, you will vote for the girl whose No. has the maximum number of unique prime factors.

For example if the girl's No. is 12, and another girl's No. is 210, then you will choose the girl with No. 210. Because 210 = 2 *3 * 5*7 , 12 = 2*2*3. 210 have 4 unique prime factors but 12 just have 2. If there are many results, you will choose the one whose name has minimum lexicographic order.

Input

The first line contain an integer T (1 <= T <= 100).Then T cases followed. Each case begins with an integer n (1 <= n <= 1000) which is the number of girls.And then followed n lines ,each line contain a string and an integer No.(1 <= No. <= 2^31 - 1). The string is the girl's name and No. is the girl's No.The string's length will not longer than 20.

Output

For each case,output the girl's name who you will vote.

Sample Input

  
   2
3
Kate 56
Lily 45
Amanda 8
4
Sara 55
Ella 42
Cristina 210
Cozzi 2

Sample Output

  
   Kate
Cristina

Source

HDU 2009-5 Programming Contest

//效率还可以，java用时984ms，由于题目没看清一个附加条件（如果因字数相同且字典顺序最小的名字）

import java.util.*;
public class hdu2574 {
	public static void main(String[] args) {	
     Scanner  sc = new Scanner(System.in);       
    	 int T=sc.nextInt();
    	 while(T-->0){
    		 int n=sc.nextInt();
    		  String s,ss = "ZZZZZZZZZZZZZZZZZZZZ";
    		 int a;
    		 int max =0;
    		 for(int i=1;i<=n;i++){
    			 s=sc.next();
    			 a=sc.nextInt();
    			 int p=prime(a);//数的不同的素数因子个数   			
    			 if(max<p||max==p&&ss.compareTo(s)>0){
    				 max=p;
    				 ss=s;   				
    			 }
    		 }System.out.println(ss);
    	 }    
	}
	private static int prime(int m) {		
		 int i=2,count=0;
           while(m>=i){
      	 
        	   if(m%i==0){
        		   count++; 
        		   while(m%i==0)
            		   m/=i;        		
        	   }   
        	    i++;        	    
           }
		return count;
	}
}