Problem:
Score : 200 points

Problem Statement

There are N people numbered 1 to N. Each person wears a red hat or a blue hat.

You are given a string s representing the colors of the people. Person i wears a red hat if s_i is R, and a blue hat if s_i is B.

Determine if there are more people wearing a red hat than people wearing a blue hat.

Constraints

1 \leq N \leq 100

|s| = N

s_i is R or B.

Input

Input is given from Standard Input in the following format:

N
s

Output

If there are more people wearing a red hat than there are people wearing a blue hat, print Yes; otherwise, print No.

Sample Input 1

4
RRBR

Sample Output 1

Yes

There are three people wearing a red hat, and one person wearing a blue hat.

Since there are more people wearing a red hat than people wearing a blue hat, the answer is Yes.

Sample Input 2

4
BRBR

Sample Output 2

No

There are two people wearing a red hat, and two people wearing a blue hat.

Since there are as many people wearing a red hat as people wearing a blue hat, the answer is No.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdlib>
#include<stack>
#include<queue>
using namespace std;

int max(int x, int y);
int min(int x, int y);
int gcd(int x, int y);	/*最大公約数*/
int lcm(int x, int y);	/*最小公倍数*/

typedef stack<int> IntStack;
typedef queue<int> IntQueue;

/*ここまでテンプレ*/



/*ここからmain関数*/

int main()
{
	int N;
	char s[100];
	cin >> N >> s;

	int r = 0;
	int b = 0;

	for (int i=0;i < N;i++)
	{
		if (s[i] == 'R')
			r++;
		else
			b++;
	}

	if (r > b)
		cout << "Yes";
	else
		cout << "No";

	return 0;
}

/*2数の大きい方を返す*/
int max(int x, int y)
{
	int max;

	if (x < y)
		max = y;
	else
		max = x;

	return max;
}

/*2数の小さい方を返す*/
int min(int x, int y)
{
	int min;

	if (x < y)
		min = x;
	else
		min = y;

	return min;
}

/*2数の最大公約数を返す*/
int gcd(int x, int y)
{
	int m;
	int gcd = 1;

	m = min(x, y);

	for (int i = 1;i <= m; i++)
	{
		if (x%i == 0 && y%i == 0)
			gcd = i;
	}

	return gcd;
}

/*2数の最小公倍数を返す*/
int lcm(int x, int y)
{
	int m;

	for (int i = 1;i <= y;i++)
	{
		if ((x*i) % y == 0) {
			m = x * i;
			break;
		}
	}

	return m;
}