C - 結構體 Homework

/*   第一題(**)定義一個結構體變量(包括年、月、日),計算該日在本年中爲第幾天?(注意考慮閏年問題),要求寫一個函數 days,實現上面的計算。 由主函數將年月日傳遞給 days 函數,計算後將日子傳遞迴主函數輸出。*/

    {
        typedef struct data{
            int year;
            int month;
            int day;
        }Data;

        Data today;
        printf("第一題:\n");
        printf("請輸入年月日:");
        scanf("%d%d%d",&today.year,&today.month,&today.day);
        printf("今天是本年中的第%d天\n",days(today));
    }

    {//子函數
        int days(Data today);
        int days(Data today){
            int monthday[]={31,28,31,30,31,30,31,31,30,31,30,31},sumDay = 0;
            for (int i = 0; i <= today.month - 1; i++) {
                sumDay += monthday[i];
            }
            if (today.month > 2) {
                if ((today.year % 4 == 0 && today.year % 100 != 0)||(today.year % 400 == 0)) {
                    sumDay += 1;
                }
            }
            return sumDay;
        }
    }

/*  第二題(***)某班有 5 個學生,三門課。分別編寫 3 個函數實現以下要求:
     (1) 求各門課的平均分;
     (2) 找出有兩門以上不及格的學生,並輸出其學號和不及格課程的成績;
     (3) 找出三門課平均成績在 85-90 分的學生,並輸出其學號和姓名*/

//  (1) 求各門課的平均分;
    //定義五個學生成績
    typedef struct score{
        float  chinese;
        float math;
        float english;
    }Score;

    typedef struct student{
        int num;
        char name[20];
        Score stuScore;
    }Student;

    typedef struct avergeM{
        float avergeC;
        float avergeM;
        float avergeE;
    }AvergeM;

    Score score1 = { 96, 98, 95};
    Score score2 = { 49, 89, 32};
    Score score3 = { 106, 48, 95};
    Score score4 = { 85, 85, 85};
    Score score5 = { 96, 98, 95};
    Student stu1 = {1, "德萊厄斯", score1};
    Student stu2 = {2, "傑斯", score2};
    Student stu3 = {3, "內瑟斯",score3 };
    Student stu4 = {4, "雷克頓", score4};
    Student stu5 = {5, "瑞文", score5};
    Student stu[] ={stu1, stu2, stu3, stu4, stu5};
    float chinese = average(stu).avergeC;
    float math = average(stu).avergeM;
    float english = average(stu).avergeE;
    printf("\n第二題(1):\n");
    printf("語文的平均分爲:%.2f\n數學的平均分爲:%.2f\n英語的平均分爲:%.2f\n",chinese,math,english);

//    地址不同
//    printf("%p\n",&score1.chinese);
//    printf("%p\n",&stu1.stuScore.chinese);
    {
        AvergeM average(Student stu[]){
            int sum3 = 0,sum1 = 0,sum2 = 0;
            AvergeM a ;
            for (int i = 0; i < 5; i++) {
                sum1 += stu[i].stuScore.chinese;
                sum2 += stu[i].stuScore.math;
                sum3 += stu[i].stuScore.english;
            }
            a.avergeC = sum1/5;
            a.avergeM = sum2/5;
            a.avergeE = sum3/5;
            return a;
        }

    }

//    (2) 找出有兩門以上不及格的學生,並輸出其學號和不及格課程的成績;
    void findStu(Student stu[]){
        printf("\n第二題:\n");int flagB = 0;
        for (int i = 0; i < 5; i++) {
            if(stu[i].stuScore.chinese < 60 && stu[i].stuScore.math < 60 && stu[i].stuScore.english < 60)
            {       printf("%d號學生兩門以上不及格.\n",stu[i].num);
                printf("語文不及格,成績爲:%d\n",stu[i].stuScore.chinese);
                printf("數學不及格,成績爲:%d\n",stu[i].stuScore.math);
                printf("英語不及格,成績爲:%d\n\n",stu[i].stuScore.chinese);
                flagB ++;
            }
        }
        if (flagB == 0) {
            printf("沒有兩門以上不及格的學生\n");
        }
    }


//(3) 找出三門課平均成績在 85-90 分的學生,並輸出其學號和姓名
    {
        void outstanding(Student stu[]){
            float averge;int flagB = 0;
            printf("\n第二題(3):\n");
            for (int i = 0; i < 5; i++) {
                averge = (stu[i].stuScore.english + stu[i].stuScore.chinese + stu[i].stuScore.math)/3;
                if ( averge >= 85 && averge <= 90) {
                    printf("學號爲%d的%s同學三門課平均成績在 85-90 分\n",stu[i].num,stu[i].name);
                    flagB ++;
                }
            }
            if (flagB == 0) {
                printf("沒有三門課平均成績在 85-90 分的學生\n\n");
            }
            else{
                printf("\n");
            }
        }
    }
/*  第三題(***)模擬n個人參加選舉的過程,並輸出選舉結果:假設候選人有四人,分別用 A、B、C、D 表示,當選某候選人時直接輸入其編號(編號由計算機隨機產生),若輸入的不是 A、B、C、D 則視爲無效票,選舉結束後按得票數從高到低輸出候選人編號和所得票數*/

    typedef struct candidate {
        char name;
        int ticket;
    }Candidate;

    Candidate per1 = {'A',0};
    Candidate per2 = {'B',0};
    Candidate per3 = {'C',0};
    Candidate per4 = {'D',0};
    Candidate person[] = {per1,per2,per3,per4};
    result(person);

    {//子函數
        void result(Candidate person[]){
            int code;
            printf("第三題:\n");
            for (int i = 0; i < 100; i++) {
                code = arc4random() % (10 - 1 + 1) + 1;
                switch (code) {
                    case 1:
                        person[0].ticket++;
                        break;
                    case 2:
                        person[1].ticket++;
                        break;
                    case 3:
                        person[2].ticket++;
                        break;
                    case 4:
                        person[3].ticket++;
                        break;
                    default:
                        break;
                }
            }
            for (int i = 0; i < 4 - 1; i++) {
                for (int j = 0; j < 4 - 1 - i; j++) {
                    if (person[j].ticket < person[j].ticket) {
                        int temp = person[j].ticket;
                        person[j].ticket = person[j + 1].ticket;
                        person[j + 1].ticket = temp;
                        char tempN = person[j].name;
                        person[j].name = person[j + 1].name;
                        person[j + 1].name = tempN;
                    }
                }
            }
            for (int k = 0; k < 4; k++) {
                printf("%c的票數爲:%d\n",person[k].name,person[k].ticket);
            }

        }
    }

//    *第四題(***)創建一個 Point 結構體,包含 x,y 兩個變量。 並寫以下函數:
//    函數 1,判斷兩個點是否在一條水平線上。
//    函數 2,判斷兩個點是否在一條垂直線上。
//    函數 3,判斷兩個點是否相等。
    {
        BOOL horizontal(PointA po1,PointA po2){
            if (po1.x == po2.x) {
                return YES;
            }
            return NO;
        }
        BOOL vertical(PointA po1,PointA po2){
            if (po1.y == po2.y) {
                return YES;
            }
            return NO;
        }
        BOOL equal(PointA po1, PointA po2){
            if (po1.x == po2.x && po1.y == po2.y) {
                return YES;
            }
            return NO;
        }
    }
    typedef struct pointA{
        float x;
        float y;
    }PointA;
    PointA po1 = {12, 56};
    PointA po2 = {12, 56};
    printf("\n第五題:\n");
    printf("兩個點 \"%s\" 在一條水平線上\n",horizontal( po1, po2) == 1 ? "是":"不");
    printf("兩個點 \"%s\" 在一條垂直線上\n",vertical( po1, po2) == 1 ? "是":"不");
    printf("兩個點 \"%s\" 相等\n",equal( po1,  po2) == 1 ? "是":"不");