首先给出自己做法,也是大多数的做法,但是最后一个测试点死活过不去,考虑到是两个for循环占用了绝大多数时间。后来再看题目时,发现分数是百分制,最后解决
#include<stdio.h>
#include<stdlib.h>
int main() {
int studentNumber;
scanf("%d", &studentNumber);
int* studentPointer = calloc(sizeof(int), studentNumber);
int i,j;
for (i = 0; i < studentNumber; i++)
scanf("%d", (studentPointer + i));
int queryNumber;
scanf("%d", &queryNumber);
int* queryPointer = calloc(sizeof(int), queryNumber);
int* countPointer = calloc(sizeof(int), queryNumber);
for (i = 0; i < queryNumber; i++)
scanf("%d", (queryPointer+i));
for (i = 0; i < studentNumber; i++) {
for (j = 0; j < queryNumber; j++) {
if (*(studentPointer + i) == *(queryPointer + j))
*(countPointer + j)+=1;
}
}
for (i = 0; i < queryNumber; i++) {
printf("%d", *(countPointer + i));
if (i != queryNumber - 1)
printf(" ");
}
return 0;
}
更正:2020年6月21日01:11:42
#include<stdio.h>
#include<stdlib.h>
int main() {
int count[101] = { 0 };
int studentNumber, i, middle, queryNumber;
scanf("%d", &studentNumber);
for (i = 0; i < studentNumber; i++) {
scanf("%d", &middle);
count[middle]++;
}
scanf("%d", &queryNumber);
int* queryPointer = calloc(sizeof(int), queryNumber);
for (i = 0; i < queryNumber; i++) {
scanf("%d", (queryPointer + i));
printf("%d", count[*(queryPointer + i)]);
if (i != queryNumber - 1)
printf(" ");
}
system("pause");
return 0;
}