第五次作业

1.教材P52 例4-8 分数转成等级 switch

#include<stdio.h>

main()

{

        int mark;

        printf("请输入学生的分数(0-100):\n");

        scanf("%d",&mark);

        switch(mark/10)

        {

        case 10:

        case 9:printf("A\n");break;

        case 8:printf("B\n");break;

        case 7:printf("C\n");break;

        case 6:printf("D\n");break;

        default:printf("NO PASS!\n");

        }

}

 

 

 

 

2.教材P58 案例三

#include<stdio.h>

main()

{

        int x,y;

        printf("请输入x的值:\n");

        scanf("%d",&x);

        if(x>0)

                 y=x*x+1;

        else if(x==0)

                 y=0;

        else

                 y=-x*x-10;

        printf("x=%d\ny=%d\n",x,y);

 

}

 

 

 

3.教材P63 案例八

#include<stdio.h>

main()

{

        float a,b,c,e,g;

        char d;

        scanf("%f",&a);

        scanf("%c",&d);

        scanf("%f",&b);

        switch(d)

        {

        case '+':

                         c=a+b;

                         printf("%.2f+%.2f=%.2f\n",a,b,c);

                                  break;

        case '-':

                 c=a-b;

                 printf("%.2f-%.2f=%.2f\n",a,b,c);

        case '*':

                 c=a*b;

                 printf("%.2f*%.2f=%.2f\n",a,b,c);

        case '/':

                 c=a/b;

                 printf("%.2f/%.2f=%.2f\n",a,b,c);

                 break;

        }

}

 

 

 

4.输入年份判断是不是闰年(闰年条件:能被4整除但不能被100整除或者能被400整除)

#include<stdio.h>

main()

{

        int a;

        printf("输入年份:");

        scanf("%d",&a);

        if(a%4==0&&a%100//=0a%400==0);

                 printf("%d年是闰年\n",a);

        else

                 printf("不是闰年\n");

 

}

 

 

 

5．编写程序，使用条件运算符找出三个数中最小的数字，并输出

#include<stdio.h>

main()

{

        float a,b,c,min;

        scanf("%f%f%f",&a,&b,&c);

        min=a<b?a:b;

        min=min<c?min:c;

        printf("Min:%f\n",min);

 

}

 

 

 

6.编写程序，判断整数m是否能被4和6同时整除

#include<stdio.h>

main()

{

        int m;

        scanf("%d",&m);

        if(m%4==0&&m%6==0)

                 printf("Yes\n");

        else

                 printf("No\n");

 

}