第五次作业

1. #include<stdio.h>

main()

{

         int mark;

         printf("输入学生的分数（0-100):\n");

         scanf("%d”,&mark);

         switch(mark/10)

         {

         case 10:              //与case9:共用一条语句

         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.编写程序，根据x的数值，求出相应y的值

#include<stdio.h>

main()

{

         float x,y;

         printf("请输入x的值\n");

         scanf("%x,&x");

         if(x>0)

                  y=x*x+1;

         else if(x==0)

                  y=0;

         else

                  y=-x*x+1;

         printf("x=%f\n=ny=%f\n",x,y);

}

3.使用多分支选择结构，实现两个数加，减，乘，除的简单计算器。

#include<stdio.h>

main()

{

         float n1,n2;

         char sign;

         printf("请输入计算的表达式:\n");

         scanf("%d%c%f",&n1,&sign,&n2);

         switch(sign)

         {

         case '+':printf("n1+n2=%f\n",n1+n2);break;

         case '-':printf("n1-n2=%f\n",n1-n2);break;

         case '*':printf("n1*n2=%f\n",n1*n2);break;

         case '/':printf("n1/n2=%f\n",n1/n2);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 n;

         printf("请输入具体年份:");

         scanf("%d",&n);

                  if(n%400==0||n%4==0&&n%100-0)

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

                  else

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

                          return 0;

 }

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");

}