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

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

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
",ans);
25     }
26     return 0;
27 }