解决要求苛刻的结构任务
Solving demanding structure task
给出结构：
struct Date {
int day, month, year;
};
struct Subject {
int id;
char name_of_subject[100];
double ects;
};
struct Grade {
char namesurname[100];
int grade_value;
struct Subject subject;
struct Date date;
};
有必要编写一个函数 digits_in_date 来接收成绩结构数组。对于所有通过的科目总计 30 个或更多 ECTS 学分的学生（他们的成绩为 6 或更高），有必要从数组中排除（不改变顺序！）那些日期包含不同数字的最小数目。例如，日期 2020 年 2 月 20 日仅包含两个不同的数字（2 和 0）。如果他们有多个成绩的日期具有相同（最小）位数，则应从数组中删除所有此类成绩。可以根据名字和姓氏查出成绩属于同一个学生（假设两个学生的名字和姓氏相同，那么就是同一个学生）。假设数据是有效的（有效日期、有效成绩、ects学分不为负等）不允许辅助数组。做直方图时只允许辅助数组。
示例：
struct Grade arr[3] = {
{"Luc Manning", 10, {1, "Physics", 25}, {19, 1, 2020}},
{"Kaia Feeney", 8, {1, "Math", 25}, {19, 1, 2020}},
{"Luc Manning", 7, {2, "Linear Algebra", 10}, {11, 1, 2020}}};
解释：
Luc Manning 的 ECTS 总分是 35，所以我们检查他的科目日期中有多少个不同的数字。 Physics 有 4 个，Linear Algebra 有 3 个不同的日期数字，所以 Linear Algebra 应该去掉。
Kaia Feeney ECTS 总分是 25，所以我们什么都不检查。
输出将是：
Luc Manning - Physics
Kaia Feeney - Math
说明：
创建一个辅助函数，使用直方图确定日期中不同数字的数量。然后是计算 ECTS 学分数量的辅助函数，最后是弹出函数。这将使您的任务变得更加容易。
我在代码中遇到的唯一问题是我的程序删除了错误的主题，而其余任务的算法是正确的。
一半的代码只是自动测试:)代码不是那么长。
编辑：
@pm100做了一个可以省400行代码的函数，加上我对原代码的修改，现在的代码是：
#include <stdio.h>
#include <string.h>
#include <sys/types.h>
struct Date {
int day, month, year;
};
struct Subject {
int id;
char name_of_subject[100];
double ects;
};
struct Grade {
char namesurname[100];
int grade_value;
struct Subject subject;
struct Date date;
};
int num_digit(int n) {
int num = 0;
while (n != 0) {
num++;
n /= 10;
}
return num;
}
int check_digits(int d, int m, int y) {
char dateStr[10];
sprintf(dateStr, "%02d%02d%04d", d, m, y);
int count[10] = {0};
for (int i = 0; i < strlen(dateStr); i++) {
char ch = dateStr[i];
count[ch - '0']++;
}
int num = 0;
for (int i = 0; i < 10; i++) {
if (count[i] > 0)
num++;
}
return num;
}
double credits(struct Grade arr[], int n, int i) {
int j;
double sum = arr[i].subject.ects;
for (j = i + 1; j < n; j++)
if (strcmp(arr[j].namesurname, arr[i].namesurname) == 0)
sum += arr[j].subject.ects;
return sum;
}
int find_min(struct Grade arr[], int n, int i) {
int j,
min = check_digits(arr[i].date.day, arr[i].date.month, arr[i].date.year);
for (j = i + 1; j < n; j++)
if (strcmp(arr[j].namesurname, arr[i].namesurname) == 0)
if (check_digits(arr[j].date.day, arr[j].date.month, arr[j].date.year) <
min)
min = j;
return min;
}
int digits_in_date(struct Grade arr[], int n) {
int i, j, k, min;
double sum_ects;
for (i = 0; i < n; i++) {
sum_ects = credits(arr, n, i);
if (sum_ects >= 30) {
min = find_min(arr, n, i);
for (j = 0; j < n; j++)
if (strcmp(arr[j].subject.name_of_subject,
arr[min].subject.name_of_subject) == 0) {
for (k = j + 1; k < n; k++)
arr[k - 1] = arr[k];
j--;
n--;
}
}
}
return n;
}
int main() {
int i, s;
printf("\nAT1: Trivial case\n");
struct Grade arr1[3] = {
{"Luc Manning", 10, {1, "Physics", 25}, {19, 1, 2020}},
{"Kaia Feeney", 8, {1, "Math", 25}, {19, 1, 2020}},
{"Luc Manning", 7, {2, "Linear Algebra", 10}, {11, 1, 2020}}};
s = digits_in_date(arr1, 3);
for (i = 0; i < s; i++)
printf("%s - %s\n", arr1[i].namesurname, arr1[i].subject.name_of_subject);
printf("\nAT2: We must not observe the same student more than once\n");
struct Grade arr2[4] = {
{"Mujo Suljic", 10, {1, "Uvod u programiranje", 25}, {19, 1, 2020}},
{"Fata Fatic", 8, {1, "Uvod u programiranje", 25}, {19, 1, 2020}},
{"Mujo Suljic", 7, {2, "Inzenjerska fizika", 10}, {11, 1, 2020}},
{"Mujo Suljic", 10, {3, "Inzenjerska matematika", 6}, {1, 9, 2020}}};
s = digits_in_date(arr2, 4);
for (i = 0; i < s; i++)
printf("%s - %s\n", arr2[i].namesurname, arr2[i].subject.name_of_subject);
printf("CORRECT:\nMujo Suljic - Uvod u programiranje\nFata Fatic - Uvod u programiranje\nMujo Suljic - Inzenjerska matematika\n");
printf("\nAT3: Multiple successive ejections\n");
struct Grade arr3[6] = {
{"Mujo Suljic", 10, {1, "Uvod u programiranje", 25}, {2, 2, 2020}},
{"Mujo Suljic", 6, {5, "Osnove elektrotehnike", 25}, {20, 2, 2022}},
{"Fata Fatic", 8, {1, "Uvod u programiranje", 25}, {19, 1, 2020}},
{"Mujo Suljic", 7, {2, "Inzenjerska fizika", 10}, {11, 1, 2020}},
{"Mujo Suljic", 10, {3, "Inzenjerska matematika", 6}, {20, 1, 2020}},
{"Fata Fatic", 8, {4, "Inzenjerska matematika", 25}, {9, 9, 2021}}};
s = digits_in_date(arr3, 6);
for (i = 0; i < s; i++)
printf("%s - %s %g\n", arr3[i].namesurname, arr3[i].subject.name_of_subject,
arr3[i].subject.ects);
printf("CORRECT:\nMujo Suljic - Inzenjerska fizika 10\nMujo Suljic - Inzenjerska matematika 6\n");
printf("\nAT4: Array with one element\n");
struct Grade arr4[1] = {
{"John Smith", 10, {1, "Subject 1", 29}, {19, 11, 1991}}};
s = digits_in_date(arr4, 1);
printf("%d ", s);
arr4[0].subject.ects = 31;
s = digits_in_date(arr4, 1);
printf("%d\n", s);
printf("CORRECT:\n1 0\n");
printf(
"\nAT5: We don't eject anything out, no one has 30 credits in total\n");
struct Grade arr5[6] = {
{"Mujo Suljic", 10, {1, "Uvod u programiranje", 25}, {20, 2, 2020}},
{"Pero Peric", 6, {5, "Osnove elektrotehnike", 25}, {20, 2, 2020}},
{"Fata Fatic", 8, {1, "Uvod u programiranje", 25}, {19, 1, 2020}},
{"Mujo Suljic", 7, {2, "Inzenjerska fizika", 1}, {11, 1, 2020}},
{"Mujo Suljic", 10, {3, "Inzenjerska matematika", 2}, {20, 1, 2020}},
{"Fata Fatic", 8, {4, "Inzenjerska matematika", 2}, {19, 1, 2020}}};
s = digits_in_date(arr5, 6);
printf("s: %d\n", s);
printf("AT5 again: Inzenjerska fizika now carries 4 ECTS credits so Mujo has "
"the required number of credits\n");
arr5[3].subject.ects = 4;
s = digits_in_date(arr5, 6);
for (i = 0; i < s; i++)
printf("%s - %s %g\n", arr5[i].namesurname, arr5[i].subject.name_of_subject,
arr5[i].subject.ects);
printf("CORRECT:\ns: 6\nAT5 again: Inzenjerska fizika now carries 4 ECTS credits so Mujo has "
"the required number of credits\nPero Peric - Osnove elektrotehnike 25\nFata Fatic - Uvod u programiranje 25\nMujo Suljic - Inzenjerska fizika 4\n"
"Mujo Suljic - Inzenjerska matematika 2\nFata Fatic - Inzenjerska matematika 2\n");
//
printf("\nAT6: Mujo Suljic did not pass Injzenjerska fizika, he does not "
"have 30 ECTS credits\n");
struct Grade arr6[6] = {
{"Mujo Suljic", 10, {1, "Uvod u programiranje", 25}, {20, 2, 2020}},
{"Pero Peric", 6, {5, "Osnove elektrotehnike", 25}, {20, 2, 2020}},
{"Fata Fatic", 8, {1, "Uvod u programiranje", 25}, {19, 1, 2020}},
{"Mujo Suljic", 5, {2, "Inzenjerska fizika", 10}, {11, 1, 2020}},
{"Mujo Suljic", 10, {3, "Inzenjerska matematika", 4}, {20, 1, 2020}},
{"Fata Fatic", 8, {4, "Inzenjerska matematika", 25}, {19, 1, 2020}}};
s = digits_in_date(arr6, 6);
for (i = 0; i < s; i++)
printf("%s - %s %g\n", arr6[i].namesurname, arr6[i].subject.name_of_subject,
arr6[i].subject.ects);
printf("CORRECT:\nMujo Suljic - Uvod u programiranje 25\nPero Peric - Osnove elektrotehnike 25\n"
"Mujo Suljic - Inzenjerska fizika 10\nMujo Suljic - Inzenjerska matematika 4\n");
return 0;
}
我只得到 AutoTest1 - AT1 的正确输出（与示例相同），其余的 AT 都是错误的。
你能帮我修改一下使其正常工作吗？
这是我的计数函数
转换为字符串，然后在 count[]
中创建分布 'histogram'
然后统计非0的个数
int count_digits(int d, int m, int y)
{
char dateStr[10];
sprintf(dateStr,"%02d%02d%04d", d, m, y);
int count[10] = { 0 };
for (int i = 0; i < strlen(dateStr); i++) {
char ch = dateStr[i];
count[ch - '0']++;
}
int tally = 0;
for (int i = 0; i < 10; i++)
{
if (count[i] > 0)
tally++;
}
return tally;
}
节省大约 380 行
我认为您误解了说明。你认为你必须拒绝一个学生除了最大的一切，不，你必须拒绝最小的。
对于 at2
{"Mujo Suljic", 10, {1, "Uvod u programiranje", 25}, {19, 1, 2020}}, <<==4
{"Fata Fatic", 8, {1, "Uvod u programiranje", 25}, {19, 1, 2020}},
{"Mujo Suljic", 7, {2, "Inzenjerska fizika", 10}, {11, 1, 2020}}, <<==3
{"Mujo Suljic", 10, {3, "Inzenjerska matematika", 6}, {1, 9, 2020}} }; <<< = 4
必须去掉第3行
{"Mujo Suljic", 7, {2, "Inzenjerska fizika", 10}, {11, 1, 2020}},
但是在你的'comment'中你说
printf("\nAT2: We must not observe the same student more than once\n");
答案显示您必须多次显示同一个学生。
我认为你可以修改代码，现在你发现你误解了赋值的逻辑
你也犯了一些地方没有从数组开头开始的错误
double credits(struct Grade arr[], int n, int i) {
int j;
double sum = arr[i].subject.ects;
for (j = i + 1; j < n; j++)
if (strcmp(arr[j].namesurname, arr[i].namesurname) == 0)
sum += arr[j].subject.ects;
return sum;
}
我加了
double credits(struct Grade arr[], int n, int i) {
int j;
double sum = arr[i].subject.ects;
for (j = i + 1; j < n; j++)
if (strcmp(arr[j].namesurname, arr[i].namesurname) == 0)
sum += arr[j].subject.ects;
printf("credits = %f for %s\n", sum, arr[i].namesurname); <<<<==
return sum;
}
对于 AT3 我得到
AT3: Multiple successive ejections
credits = 66.000000 for Mujo Suljic
credits = 25.000000 for Fata Fatic
不正确
法塔有 50 个。
这是因为你开始了一半。我认为你在最小计算中犯了同样的错误
最后你有一些非常简单的错误
我留给你找出问题所在（暗示你对 'min' 是什么感到困惑，加上其他 2 个错误）
int find_min(struct Grade arr[], int n, int i) {
int j,
min = 10;// check_digits(arr[i].date.day, arr[i].date.month, arr[i].date.year);
int minIdx = -1;
for (j = i ; j < n; j++)
if (strcmp(arr[j].namesurname, arr[i].namesurname) == 0)
if (check_digits(arr[j].date.day, arr[j].date.month, arr[j].date.year) <
min)
minIdx = j;
return minIdx;
}
给出结构：
struct Date {
int day, month, year;
};
struct Subject {
int id;
char name_of_subject[100];
double ects;
};
struct Grade {
char namesurname[100];
int grade_value;
struct Subject subject;
struct Date date;
};
有必要编写一个函数 digits_in_date 来接收成绩结构数组。对于所有通过的科目总计 30 个或更多 ECTS 学分的学生（他们的成绩为 6 或更高），有必要从数组中排除（不改变顺序！）那些日期包含不同数字的最小数目。例如，日期 2020 年 2 月 20 日仅包含两个不同的数字（2 和 0）。如果他们有多个成绩的日期具有相同（最小）位数，则应从数组中删除所有此类成绩。可以根据名字和姓氏查出成绩属于同一个学生（假设两个学生的名字和姓氏相同，那么就是同一个学生）。假设数据是有效的（有效日期、有效成绩、ects学分不为负等）不允许辅助数组。做直方图时只允许辅助数组。
示例：
struct Grade arr[3] = {
{"Luc Manning", 10, {1, "Physics", 25}, {19, 1, 2020}},
{"Kaia Feeney", 8, {1, "Math", 25}, {19, 1, 2020}},
{"Luc Manning", 7, {2, "Linear Algebra", 10}, {11, 1, 2020}}};
解释：
Luc Manning 的 ECTS 总分是 35，所以我们检查他的科目日期中有多少个不同的数字。 Physics 有 4 个，Linear Algebra 有 3 个不同的日期数字，所以 Linear Algebra 应该去掉。 Kaia Feeney ECTS 总分是 25，所以我们什么都不检查。
输出将是：
Luc Manning - Physics
Kaia Feeney - Math
说明： 创建一个辅助函数，使用直方图确定日期中不同数字的数量。然后是计算 ECTS 学分数量的辅助函数，最后是弹出函数。这将使您的任务变得更加容易。
我在代码中遇到的唯一问题是我的程序删除了错误的主题，而其余任务的算法是正确的。 一半的代码只是自动测试:)代码不是那么长。
编辑： @pm100做了一个可以省400行代码的函数，加上我对原代码的修改，现在的代码是：
#include <stdio.h>
#include <string.h>
#include <sys/types.h>
struct Date {
int day, month, year;
};
struct Subject {
int id;
char name_of_subject[100];
double ects;
};
struct Grade {
char namesurname[100];
int grade_value;
struct Subject subject;
struct Date date;
};
int num_digit(int n) {
int num = 0;
while (n != 0) {
num++;
n /= 10;
}
return num;
}
int check_digits(int d, int m, int y) {
char dateStr[10];
sprintf(dateStr, "%02d%02d%04d", d, m, y);
int count[10] = {0};
for (int i = 0; i < strlen(dateStr); i++) {
char ch = dateStr[i];
count[ch - '0']++;
}
int num = 0;
for (int i = 0; i < 10; i++) {
if (count[i] > 0)
num++;
}
return num;
}
double credits(struct Grade arr[], int n, int i) {
int j;
double sum = arr[i].subject.ects;
for (j = i + 1; j < n; j++)
if (strcmp(arr[j].namesurname, arr[i].namesurname) == 0)
sum += arr[j].subject.ects;
return sum;
}
int find_min(struct Grade arr[], int n, int i) {
int j,
min = check_digits(arr[i].date.day, arr[i].date.month, arr[i].date.year);
for (j = i + 1; j < n; j++)
if (strcmp(arr[j].namesurname, arr[i].namesurname) == 0)
if (check_digits(arr[j].date.day, arr[j].date.month, arr[j].date.year) <
min)
min = j;
return min;
}
int digits_in_date(struct Grade arr[], int n) {
int i, j, k, min;
double sum_ects;
for (i = 0; i < n; i++) {
sum_ects = credits(arr, n, i);
if (sum_ects >= 30) {
min = find_min(arr, n, i);
for (j = 0; j < n; j++)
if (strcmp(arr[j].subject.name_of_subject,
arr[min].subject.name_of_subject) == 0) {
for (k = j + 1; k < n; k++)
arr[k - 1] = arr[k];
j--;
n--;
}
}
}
return n;
}
int main() {
int i, s;
printf("\nAT1: Trivial case\n");
struct Grade arr1[3] = {
{"Luc Manning", 10, {1, "Physics", 25}, {19, 1, 2020}},
{"Kaia Feeney", 8, {1, "Math", 25}, {19, 1, 2020}},
{"Luc Manning", 7, {2, "Linear Algebra", 10}, {11, 1, 2020}}};
s = digits_in_date(arr1, 3);
for (i = 0; i < s; i++)
printf("%s - %s\n", arr1[i].namesurname, arr1[i].subject.name_of_subject);
printf("\nAT2: We must not observe the same student more than once\n");
struct Grade arr2[4] = {
{"Mujo Suljic", 10, {1, "Uvod u programiranje", 25}, {19, 1, 2020}},
{"Fata Fatic", 8, {1, "Uvod u programiranje", 25}, {19, 1, 2020}},
{"Mujo Suljic", 7, {2, "Inzenjerska fizika", 10}, {11, 1, 2020}},
{"Mujo Suljic", 10, {3, "Inzenjerska matematika", 6}, {1, 9, 2020}}};
s = digits_in_date(arr2, 4);
for (i = 0; i < s; i++)
printf("%s - %s\n", arr2[i].namesurname, arr2[i].subject.name_of_subject);
printf("CORRECT:\nMujo Suljic - Uvod u programiranje\nFata Fatic - Uvod u programiranje\nMujo Suljic - Inzenjerska matematika\n");
printf("\nAT3: Multiple successive ejections\n");
struct Grade arr3[6] = {
{"Mujo Suljic", 10, {1, "Uvod u programiranje", 25}, {2, 2, 2020}},
{"Mujo Suljic", 6, {5, "Osnove elektrotehnike", 25}, {20, 2, 2022}},
{"Fata Fatic", 8, {1, "Uvod u programiranje", 25}, {19, 1, 2020}},
{"Mujo Suljic", 7, {2, "Inzenjerska fizika", 10}, {11, 1, 2020}},
{"Mujo Suljic", 10, {3, "Inzenjerska matematika", 6}, {20, 1, 2020}},
{"Fata Fatic", 8, {4, "Inzenjerska matematika", 25}, {9, 9, 2021}}};
s = digits_in_date(arr3, 6);
for (i = 0; i < s; i++)
printf("%s - %s %g\n", arr3[i].namesurname, arr3[i].subject.name_of_subject,
arr3[i].subject.ects);
printf("CORRECT:\nMujo Suljic - Inzenjerska fizika 10\nMujo Suljic - Inzenjerska matematika 6\n");
printf("\nAT4: Array with one element\n");
struct Grade arr4[1] = {
{"John Smith", 10, {1, "Subject 1", 29}, {19, 11, 1991}}};
s = digits_in_date(arr4, 1);
printf("%d ", s);
arr4[0].subject.ects = 31;
s = digits_in_date(arr4, 1);
printf("%d\n", s);
printf("CORRECT:\n1 0\n");
printf(
"\nAT5: We don't eject anything out, no one has 30 credits in total\n");
struct Grade arr5[6] = {
{"Mujo Suljic", 10, {1, "Uvod u programiranje", 25}, {20, 2, 2020}},
{"Pero Peric", 6, {5, "Osnove elektrotehnike", 25}, {20, 2, 2020}},
{"Fata Fatic", 8, {1, "Uvod u programiranje", 25}, {19, 1, 2020}},
{"Mujo Suljic", 7, {2, "Inzenjerska fizika", 1}, {11, 1, 2020}},
{"Mujo Suljic", 10, {3, "Inzenjerska matematika", 2}, {20, 1, 2020}},
{"Fata Fatic", 8, {4, "Inzenjerska matematika", 2}, {19, 1, 2020}}};
s = digits_in_date(arr5, 6);
printf("s: %d\n", s);
printf("AT5 again: Inzenjerska fizika now carries 4 ECTS credits so Mujo has "
"the required number of credits\n");
arr5[3].subject.ects = 4;
s = digits_in_date(arr5, 6);
for (i = 0; i < s; i++)
printf("%s - %s %g\n", arr5[i].namesurname, arr5[i].subject.name_of_subject,
arr5[i].subject.ects);
printf("CORRECT:\ns: 6\nAT5 again: Inzenjerska fizika now carries 4 ECTS credits so Mujo has "
"the required number of credits\nPero Peric - Osnove elektrotehnike 25\nFata Fatic - Uvod u programiranje 25\nMujo Suljic - Inzenjerska fizika 4\n"
"Mujo Suljic - Inzenjerska matematika 2\nFata Fatic - Inzenjerska matematika 2\n");
//
printf("\nAT6: Mujo Suljic did not pass Injzenjerska fizika, he does not "
"have 30 ECTS credits\n");
struct Grade arr6[6] = {
{"Mujo Suljic", 10, {1, "Uvod u programiranje", 25}, {20, 2, 2020}},
{"Pero Peric", 6, {5, "Osnove elektrotehnike", 25}, {20, 2, 2020}},
{"Fata Fatic", 8, {1, "Uvod u programiranje", 25}, {19, 1, 2020}},
{"Mujo Suljic", 5, {2, "Inzenjerska fizika", 10}, {11, 1, 2020}},
{"Mujo Suljic", 10, {3, "Inzenjerska matematika", 4}, {20, 1, 2020}},
{"Fata Fatic", 8, {4, "Inzenjerska matematika", 25}, {19, 1, 2020}}};
s = digits_in_date(arr6, 6);
for (i = 0; i < s; i++)
printf("%s - %s %g\n", arr6[i].namesurname, arr6[i].subject.name_of_subject,
arr6[i].subject.ects);
printf("CORRECT:\nMujo Suljic - Uvod u programiranje 25\nPero Peric - Osnove elektrotehnike 25\n"
"Mujo Suljic - Inzenjerska fizika 10\nMujo Suljic - Inzenjerska matematika 4\n");
return 0;
}
我只得到 AutoTest1 - AT1 的正确输出（与示例相同），其余的 AT 都是错误的。 你能帮我修改一下使其正常工作吗？
这是我的计数函数
转换为字符串，然后在 count[]
中创建分布 'histogram'然后统计非0的个数
int count_digits(int d, int m, int y)
{
char dateStr[10];
sprintf(dateStr,"%02d%02d%04d", d, m, y);
int count[10] = { 0 };
for (int i = 0; i < strlen(dateStr); i++) {
char ch = dateStr[i];
count[ch - '0']++;
}
int tally = 0;
for (int i = 0; i < 10; i++)
{
if (count[i] > 0)
tally++;
}
return tally;
}
节省大约 380 行
我认为您误解了说明。你认为你必须拒绝一个学生除了最大的一切，不，你必须拒绝最小的。
对于 at2
{"Mujo Suljic", 10, {1, "Uvod u programiranje", 25}, {19, 1, 2020}}, <<==4
{"Fata Fatic", 8, {1, "Uvod u programiranje", 25}, {19, 1, 2020}},
{"Mujo Suljic", 7, {2, "Inzenjerska fizika", 10}, {11, 1, 2020}}, <<==3
{"Mujo Suljic", 10, {3, "Inzenjerska matematika", 6}, {1, 9, 2020}} }; <<< = 4
必须去掉第3行
{"Mujo Suljic", 7, {2, "Inzenjerska fizika", 10}, {11, 1, 2020}},
但是在你的'comment'中你说
printf("\nAT2: We must not observe the same student more than once\n");
答案显示您必须多次显示同一个学生。
我认为你可以修改代码，现在你发现你误解了赋值的逻辑
你也犯了一些地方没有从数组开头开始的错误
double credits(struct Grade arr[], int n, int i) {
int j;
double sum = arr[i].subject.ects;
for (j = i + 1; j < n; j++)
if (strcmp(arr[j].namesurname, arr[i].namesurname) == 0)
sum += arr[j].subject.ects;
return sum;
}
我加了
double credits(struct Grade arr[], int n, int i) {
int j;
double sum = arr[i].subject.ects;
for (j = i + 1; j < n; j++)
if (strcmp(arr[j].namesurname, arr[i].namesurname) == 0)
sum += arr[j].subject.ects;
printf("credits = %f for %s\n", sum, arr[i].namesurname); <<<<==
return sum;
}
对于 AT3 我得到
AT3: Multiple successive ejections
credits = 66.000000 for Mujo Suljic
credits = 25.000000 for Fata Fatic
不正确
法塔有 50 个。
这是因为你开始了一半。我认为你在最小计算中犯了同样的错误
最后你有一些非常简单的错误
我留给你找出问题所在（暗示你对 'min' 是什么感到困惑，加上其他 2 个错误）
int find_min(struct Grade arr[], int n, int i) {
int j,
min = 10;// check_digits(arr[i].date.day, arr[i].date.month, arr[i].date.year);
int minIdx = -1;
for (j = i ; j < n; j++)
if (strcmp(arr[j].namesurname, arr[i].namesurname) == 0)
if (check_digits(arr[j].date.day, arr[j].date.month, arr[j].date.year) <
min)
minIdx = j;
return minIdx;
}