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.
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.
For each case,output the girl‘s name who you will vote.
2
3
Kate 56
Lily 45
Amanda 8
4
Sara 55
Ella 42
Cristina 210
Cozzi 2
#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <iostream>
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define LL __int64
#define PI 3.1415926
const int maxn = 1000000;
int a[maxn];
struct node
{
char name[2222];
int score;
int sum;
};
node f[1005];
bool cmp(node aa,node bb)
{
if(aa.sum==bb.sum)
return strcmp(aa.name,bb.name)<0;
else
return aa.sum>bb.sum;
}
int main()
{
int t,ans;
cin>>t;
while(t--)
{
int n;
cin>>n;
int i,j;
int max;
max=-1;
for(i=0; i<n; i++)
{
ans=0;
cin>>f[i].name>>f[i].score;
for(j=2; j*j<=f[i].score; j++)
{
if(f[i].score&&f[i].score%j==0)
{
ans++;
while(f[i].score&&f[i].score%j==0)
f[i].score/=j;
}
}
if(f[i].score>1)
ans++;
f[i].sum=ans;
}
sort(f,f+n,cmp);
cout<<f[0].name<<endl;
}
return 0;
}