20200229 training race L2 - 2 tree species Statistics (25 points)

With the application of satellite imaging technology , The natural resources research institute can identify the type of each tree . Please write a program to help researchers count the number of each tree , Calculate the percentage of each tree in the total .

Input format :

The input first gives a positive integer N（≤10^5）, And then N That's ok , Each line gives the species name of a tree observed by the satellite . The category name consists of no more than 30 English letters and spaces （ Case insensitive ）.

Output format :

Incrementally output the category names of various trees and their percentage in the total number in dictionary order , Separated by spaces , After decimal point 4 position .

sample input :

29
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow

sample output :

Ash 13.7931%
Aspen 3.4483%
Basswood 3.4483%
Beech 3.4483%
Black Walnut 3.4483%
Cherry 3.4483%
Cottonwood 3.4483%
Cypress 3.4483%
Gum 3.4483%
Hackberry 3.4483%
Hard Maple 3.4483%
Hickory 3.4483%
Pecan 3.4483%
Poplan 3.4483%
Red Alder 3.4483%
Red Elm 3.4483%
Red Oak 6.8966%
Sassafras 3.4483%
Soft Maple 3.4483%
Sycamore 3.4483%
White Oak 10.3448%
Willow 3.4483%
Yellow Birch 3.4483%

map Key value pair set , key The only one is not repeated and orderly （ From small to large ）, Yes find() Function lookup key , If it cannot be found, it equals end()
find ：m.find(k)!=m.end()
Can't find ：m.find(k2)==m.end()

#include <bits/stdc++.h>
using namespace std;
int main()
{
    
	int n;
	cin>>n;
	getchar();
	map<string,int>m;
	for(int i=0;i<n;i++)
	{
    
		string s;
		getline(cin,s);
		m[s]++;// The value can also be taken directly m[k];
	}
	for(auto i=m.begin();i!=m.end();i++)
	{
    //auto stay c++11, Need configuration dev, Can also be for(map<string,int>::iterator i=m.begin();i!=m.end();i++)
		cout<<i->first<<" ";//i->first key 
		printf("%.4lf%%\n",i->second*1.0/n*100);//i->second value 
	}
	return 0;
}