• ios学习记录 day2


    1.&&和||的短路现象

    2.条件 ? 表达式1 :表达式2 ;(条件为真时执行表达式1,否则执行表达式2)

    作业:

    1.铁路托运行李规定:行李重不超过50公斤的托运费按0.15元每公斤计算,超过50公斤的部分每公斤加收0.1元。

    int main(int argc, const char * argv[])
    {

        float weight = 0,money = 0;
        printf("输入货物重量:");
        scanf("%f",&weight);
        if (weight <= 50) {
            money = weight * 0.15;
        }else if (weight > 50){
            money = (weight - 50) * 0.25 + 50 * 0.15;
        }
        printf("托运费为:%.2f ",money);

    return 0;
    }

    2.完成四则运算,如:用户输入34+56 输出90.00。用户输入两个数和运算符。通过运算符输出结果。

    (1)

    int main(int argc, const char * argv[])
    {

        float result = 0,a = 0,b = 0;
        char c = '';
        printf("请输入表达式:");
        scanf("%f%c%f",&a,&c,&b);
        if (c == '+') {
            result = a + b;
        }else if (c == '-'){
            result = a - b;
        }else if (c == '*'){
            result = a * b;
        }else if (c == '/'){
            result = a / b;
        }else {
        printf("输入错误 ");
        }
        printf("结果是:%.2f ",result);

    return 0;
    }

    (2)

    int main(int argc, const char * argv[])
    {

        float a = 0,b = 0;
        char c = '';
        printf("输入计算内容:");
        scanf("%f%c%f",&a,&c,&b);
        switch (c) {
            case '+':
                printf("结果是:%.2f ",a + b);
                break;
            case '-':
                printf("结果是:%.2f ",a - b);
                break;
            case '*':
                printf("结果是:%.2f ",a * b);
                break;
            case '/':
                printf("结果是:%.2f ",a / b);
                break;
            default:
                printf("输入错误 ");
                break;
        }

    return 0;
    }

    3.输入三个数,求最大值和中间值。

    (1)

    int main(int argc, const char * argv[])
    {

        int a = 0,b = 0,c = 0,max = 0,mid = 0;
        scanf("%d,%d,%d",&a,&b,&c);
        max = a > b ? a : b;
        max = max > c ? max : c;
        printf("最大值是:%d ",max);
        if ((a - c) * (a - b) <= 0) {
            mid = a;
        }else if ((b - a) * (b - c) <= 0){
            mid = b;
        }else{
            mid = c;
        }
        printf("中间值是:%d ",mid);

    (2)

        int a = 0,b = 0,c = 0,mid = 0;
        printf("请输入三个数:");
        scanf("%d,%d,%d",&a,&b,&c);
        if (a >= b) {
            if (a >= c) {
                (b >= c) ? (mid = b) : (mid = c);
            }else if (a < c){

               mid = a;
            }
        }else if(a < b){
            if (a >= c) {
                mid = a;
            }else if(a < c){
                (b >= c) ? (mid = c) : (mid = b);
            }
        }
        printf("中间数是:%d ",mid);

    (3)
        int a = 0,b = 0,c = 0,mid = 0;
        printf("请输入三个数:");
        scanf("%d,%d,%d",&a,&b,&c);
        if (a > b && a > c) {
            mid = (b > c) ? b : c;
        }else if (b > c && b > a){
            mid = (a > c) ? a : c;
        }else{
            mid = (a > b) ? a : b;
        }
        printf("中间值是:%d ",mid);

    (4)

        int a = 0,b = 0,c = 0,mid = 0;
        scanf("%d,%d,%d",&a,&b,&c);
        if (a >= b && a <= c) {
            mid = a;
        }else if (b >= a && b <= c){
            mid = b;
        }else{
            mid = c;
        }
        printf("中间值是:%d ",mid);

    (5)

        int a = 0,b = 0,c = 0,mid = 0,max = 0,min = 0,sum = 0;
        scanf("%d,%d,%d",&a,&b,&c);
        max = a > b ? a : b;
        max = max > c ? max : c;
        min = a < b ? a : b;
        min = min < c ? min : c;
        sum = a + b + c;
        mid = sum - max - min;
        printf("中间值是:%d ",mid);

    return 0;
    }

    4.有一个函数:x<1时,y=x;1<=x<10时,y=2x-1;x>=10时,y=3x-11;输入x,输出y。

    int main(int argc, const char * argv[])
    {

        int x = 0,y = 0;
        scanf("%d",&x);
        if (x < 1) {
            y = x;
        }else if (x >= 1 && x <= 10){
            y = 2 * x - 1;
        }else if(x > 10){
            y = 3 * x - 11;
        }
        printf("y的值是:%d ",y);

    return 0;
    }

    5.给出一个百分制成绩,90分以上为‘A’,80~89为‘B’,70~79为‘C’,60~69为‘D’,60以下为‘E’。

    int main(int argc, const char * argv[])
    {

        int x = 0;


        scanf("%d",&x);


        if (x >= 90 && x <= 100) {
            printf("A ");
        }else if (x >= 80 && x <= 89){
            printf("B ");
        }else if (x >= 70 && x <= 79){
            printf("C ");
        }else if (x >= 60 && x <= 69){
            printf("D ");
        }else if (x >= 0 && x <= 59){
            printf("E ");
        }else{
            printf("输入错误 ");
        }

    return 0;
    }

    6.输入一个五位数,判断是否为回文数(如12321)

    int main(int argc, const char * argv[])
    {  int a = 0,wuwei = 0,siwei = 0,sanwei = 0,erwei = 0,yiwei = 0;
        printf("输入一个五位数:");
        scanf("%d",&a);
        wuwei = a / 10000;
        siwei = (a - wuwei * 10000) / 1000;
        sanwei = (a - wuwei * 10000 - siwei * 1000) / 100;
        erwei = (a - wuwei * 10000 - siwei * 1000- sanwei * 100) / 10;
        yiwei = a - wuwei * 10000 - siwei * 1000- sanwei * 100 - erwei * 10;
        if (wuwei == yiwei && siwei == erwei) {
            printf("这个五位数是回文数 ");
        }else{
            printf("这个五位数不是回文数 ");
        }// 第一种

        int a = 0;
        scanf("%d",&a);
        if ((a / 10000 == a % 10) && (a / 1000) % 10 == (a % 100) / 10) {
            printf("是回文数 ");
        }else{
            printf("不是回文数 ");
        }//第二种
        return 0;
    }

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  • 原文地址:https://www.cnblogs.com/lxllanou/p/3576827.html
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