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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>

int main(void)
{
    int N;
    int Red_count = 0;
    int Blue_count = 0;    
    scanf("%d", &N);
    char s[N], str[1];
    scanf("%s", s);

    for(int i = 0; i < N; i++){
        strncpy(str, s+i, 1);

        if(strcmp("R", str) == 0){
            Red_count = Red_count + 1;
        }
        else if (strcmp("B", str) == 0){
            Blue_count = Blue_count + 1;
        }        
        
    }
    /* printf("Red_count : %d\n", Red_count); */
    /* printf("Blue_count : %d\n", Blue_count);     */
    if( Red_count > Blue_count ){
        printf("Yes\n");        
    }
    else{
        printf("No\n");
    }
    
    
    return 0;    
}