本题要求读入 N 名学生的成绩，将获得某一给定分数的学生人数输出。

输入格式：

输入在第 1 行给出不超过 105 的正整数 N，即学生总人数。随后一行给出 N 名学生的百分制整数成绩，中间以空格分隔。最后一行给出要查询的分数个数 K（不超过 N 的正整数），随后是 K 个分数，中间以空格分隔。

输出格式：

在一行中按查询顺序给出得分等于指定分数的学生人数，中间以空格分隔，但行末不得有多余空格。

输入样例：

10
60 75 90 55 75 99 82 90 75 50
3 75 90 88

输出样例：

3 2 0

编程语言：Python、C

解题思想（Python）：

1.Python直接使用count方法进行查询

但使用Python会导致最后一个测试点超时

一开始认为超时是因为Python性能的缘故，因此尝试了Java、C语言使用双层循环计数的方法，但是仍然超时

解题思想（C）：

1.突然想到一个办法，何不直接创建一个100长度的数组，每输入一个成绩就将对应下标的元素值+1

这样就省去了查询的时间，直接输出查询的成绩对应下标的数组元素即可，使用这个方法大大提高了效率！

时间复杂度：两个算法均O(N)

代码如下（Python）：

NOS = int(input())  # 输入成绩个数,Number Of Students
Score = list(map(int, input().split()))  # 存放学生成绩
Query = list(map(int, input().split()))  # 存放查询次数、具体查询的成绩
NOQ = Query[0]  # 将查询次数单独存放，便于后续操作，Number Of Queries
Query.pop(0)  # 将查询次数从列表从删除，只保留具体查询的成绩，便于后续操作
 
for i in range(NOQ - 1):  # 为了保证输出没有多余的空格，只查询n-1个成绩
    print(Score.count(Query[i]), end=' ')
print(Score.count(Query[len(Query) - 1]))  # 查询第n个成绩

代码如下（C）：

#include<stdio.h>
 
int main(void){
    int Frequency[100];//创建一个数组，存放0-100分每个成绩的个数
    for(int i=0;i<100;i++)
    {
        Frequency[i]=0;//数组初始化，将每个成绩的个数默认为0
    }
	int NOS;//成绩个数
    scanf("%d",&NOS);
	int Score;
	for(int i=0;i<NOS;i++)
	{
		scanf("%d",&Score);//输入成绩
        Frequency[Score]++;//将成绩对应下标的数组元素+1，例如76分，则Frequency[76]++
	}
	int NOQ;//查询成绩的个数
    scanf("%d",&NOQ);
	int Query[NOQ];
	for(int i=0;i<NOQ;i++)
	{
		scanf("%d",&Query[i]);//输入查询的成绩
	}
	for(int i=0;i<NOQ-1;i++)//输出前n-1个成绩对应的次数
	{
		printf("%d ",Frequency[Query[i]]);
	}
	printf("%d",Frequency[Query[NOQ-1]]);//输出第n个成绩对应的次数
}

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PAT乙级:1038 统计同成绩学生