zoukankan      html  css  js  c++  java
  • HDOJ 2574 Hdu Girls' Day

    Hdu Girls' Day

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 742    Accepted Submission(s): 241


    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
     1 #include<stdio.h>
     2 #include<string.h>
     3 #include<math.h>
     4 int prime[1000000] = {0};
     5 
     6 int main()
     7 {
     8     int T,i,j,n,num,max,k;
     9     char name[100],s[100];
    10     prime[0]=prime[1]=1;
    11     for(i = 2; i <1000; i++)
    12         for(j = i * i; j <1000000; j+= i)
    13             prime[j] = 1;
    14     scanf("%d",&T);
    15     while(T--)
    16     {
    17         max=-1;
    18         scanf("%d",&n);
    19         while(n--)
    20         {
    21             k=0;
    22             scanf("%s%d",name,&num);
    23             for(i=2;i<=num;++i)
    24             {
    25                 if(prime[i])
    26                     continue;
    27                 if(num%i==0)
    28                 {
    29                     ++k;
    30                     num/=i;
    31                 }
    32             }
    33             if(max<k)
    34             {
    35                 max=k;
    36                 strcpy(s,name);
    37             }
    38             else if(max&&max==k)
    39             {
    40                 if(strcmp(s,name)>0)
    41                     strcpy(s,name);
    42             }
    43         }
    44         printf("%s\n",s);
    45     }
    46     return 0;
    47 }
    功不成,身已退
  • 相关阅读:
    PaaS 7层动态路由的若干实现
    05-OC对象的内存分析
    04-类与对象的练习(第二个OC的类)
    03-类的声明和实现(第一个OC的类)
    02-类与对象的关系
    01-面向对象和面向过程
    06-BOOL类型的使用
    05-初识OC多文件编程(第4个OC程序)
    04-初识OC多文件编程(第3个OC程序)
    03-第二个OC程序
  • 原文地址:https://www.cnblogs.com/dongsheng/p/2636920.html
Copyright © 2011-2022 走看看