L1-023 output gplt (20 points)

L1-023 Output GPLT (20 branch )

A given length does not exceed 10000 Of 、 A string consisting only of English letters . Please reorder the characters , Press GPLTGPLT.... Output in this order , And ignore the other characters . Of course , Four types of characters （ Case insensitive ） It doesn't have to be the same number , If a character has been printed out , Then press the remaining characters GPLT Sequential printing , Until all characters are printed .

Input format ：
The input is given in a line with a length not exceeding 10000 Of 、 A non - empty string consisting only of English letters .

Output format ：
Output the sorted string on a line by the title . The question guarantees that the output is not empty .

sample input ：

pcTclnGloRgLrtLhgljkLhGFauPewSKgt

sample output ：

GPLTGPLTGLTGLGLL

First, convert all the characters in the string into uppercase letters , Then count the number of corresponding letters , Just keep outputting

tolower() The function converts all strings into lowercase letters
toupper() The function converts all strings into uppercase letters

AC Code

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<stack>
#include<queue>
#include<sstream>

using namespace std;
typedef long long ll;
const int N=100010;

int main()
{
    
	int G = 0, P = 0, L = 0, T = 0;
	string a;
	cin >> a;
	int len = a.size();
	for(int i = 0; i < len; i ++ )
	{
    
		if(toupper(a[i]) == 'G') G ++ ;
		else if(toupper(a[i]) == 'P') P ++ ;
		else if(toupper(a[i]) == 'L') L ++ ;
		else if(toupper(a[i]) == 'T') T ++ ;
	}
	while(G || P || L || T)
	{
    
		if(G)
		{
    
			cout << 'G';
			G -- ;
		}
		if(P)
		{
    
			cout << 'P';
			P -- ;
		}
		if(L)
		{
    
			cout << 'L';
			L -- ;
		}
		if(T)
		{
    
			cout << 'T';
			T -- ;
		}
	}
	return 0;
}