[HackerRank]Apple and Orange

문제

Sam's house has an apple tree and an orange tree that yield an abundance of fruit. In the diagram below, the red region denotes his house, where s is the start point, and t is the endpoint. The apple tree is to the left of his house, and the orange tree is to its right. You can assume the trees are located on a single point, where the apple tree is at point a, and the orange tree is at point b.

Apple and orange(2).png

When a fruit falls from its tree, it lands d units of distance from its tree of origin along the x-axis. A negative value of d means the fruit fell d units to the tree's left, and a positive value of d means it falls d units to the tree's right.

Given the value of d for m apples and n oranges, determine how many apples and oranges will fall on Sam's house (i.e., in the inclusive range [s,t])?

Function Description
Complete the countApplesAndOranges function in the editor below. It should print the number of apples and oranges that land on Sam's house, each on a separate line.
Input Format
The first line contains two space-separated integers denoting the respective values of s and t.
The second line contains two space-separated integers denoting the respective values of a and b.
The third line contains two space-separated integers denoting the respective values of m and n.
The fourth line contains m space-separated integers denoting the respective distances that each apple falls from point a.
The fifth line contains n space-separated integers denoting the respective distances that each orange falls from point b.
Output Format
Print two integers on two different lines:
1. The first integer: the number of apples that fall on Sam's house.
2. The second integer: the number of oranges that fall on Sam's house.

CODE

void countApplesAndOranges(int s, int t, int a, int b, int apples_count, int* apples, int oranges_count, int* oranges) {
    int count1=0, count2 = 0;
    for (int i=0;i<apples_count;i++) // 집 안에 떨어진 사과의 개수 세기
        if ((apples[i]>=(s-a))&&(apples[i]<=(t-a))) count1++; // 범위 안에 있으면 count 증가
    printf("%d\n", count1); // 출력
    for (int i=0;i<oranges_count;i++) // 집 안에 떨어진 오렌지의 개수 세기
        if ((oranges[i]>=(s-b))&&(oranges[i]<=(t-b))) count2++; // 범위 안에 있으면 count 증가
    printf("%d\n", count2); // 출력
}

감상

count 하는 변수를 하나만 쓰기 위해서 한 번 개수를 세고 for문 int i 옆에 count=0으로 초기화해서 풀려고 했다. case 0, 1, 2는 통과했지만 그 외의 case는 실패했다. 왤까?