Codeforces Gym101502 E.The Architect Omar-find()函数

E. The Architect Omar

 

time limit per test

1.0 s

memory limit per test

256 MB

input

standard input

output

standard output

Architect Omar is responsible for furnishing the new apartments after completion of its construction. Omar has a set of living room furniture, a set of kitchen furniture, and a set of bedroom furniture, from different manufacturers.

In order to furnish an apartment, Omar needs a living room furniture, a kitchen furniture, and two bedroom furniture, regardless the manufacturer company.

You are given a list of furniture Omar owns, your task is to find the maximum number of apartments that can be furnished by Omar.

Input

The first line contains an integer T (1 ≤ T ≤ 100), where T is the number of test cases.

The first line of each test case contains an integer n (1 ≤ n ≤ 1000), where n is the number of available furniture from all types. Then nlines follow, each line contains a string s representing the name of a furniture.

Each string s begins with the furniture's type, then followed by the manufacturer's name. The furniture's type can be:

• bed, which means that the furniture's type is bedroom.
• kitchen, which means that the furniture's type is kitchen.
• living, which means that the furniture's type is living room.

All strings are non-empty consisting of lowercase and uppercase English letters, and digits. The length of each of these strings does not exceed 50 characters.

Output

For each test case, print a single integer that represents the maximum number of apartments that can be furnished by Omar

Example

input

1
6
bedXs
kitchenSS1
kitchen2
bedXs
living12
livingh

output

1

这个题水题，用find()写。

代码:

 1 //E. The Architect Omar-find函数
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<queue>
 6 #include<algorithm>
 7 #include<cmath>
 8 using namespace std;
 9 int main(){
10     int t,n;
11     scanf("%d",&t);
12     while(t--){
13         scanf("%d",&n);
14         int num1=0,num2=0,num3=0;
15         for(int i=0;i<n;i++){
16             string s;
17             cin>>s;
18             if(s.find("bed")==0)num1++;
19             if(s.find("kitchen")==0)num2++;
20             if(s.find("living")==0)num3++;
21         }
22         //cout<<num1<<" "<<num2<<" "<<num3<<endl;
23         int ans=min(num1/2,min(num2,num3));
24         printf("%d\n",ans);
25     }
26     return 0;
27 }