switch case与while语句练习

switch case 选择

• 假设用1，2。。。。7分别表示星期一。。。。星期天，现输入一个数字，输出对应的星期几。比如：输入3，则输出“星期三”

  #define _CRT_SECURE_NO_WARNINGS 1
  #include 
  int main()
  {
  	int day = 0;
  	scanf("%d", &day);
  	switch (day)
  	{
  		case 1:
  			printf("星期一\n");
  			break;
  		case 2:
  			printf("星期二\n");
  			break;
  		case 3:
  			printf("星期三\n");
  			break;
  		case 4:
  			printf("星期四\n");
  			break;
  		case 5:
  			printf("星期五\n");
  			break;
  		case 6:
  			printf("星期六\n");
  			break;
  		case 7:
  			printf("星期天\n");
  			break;
  	}
  	return 0;
  }

• ---

  从键盘上输入学生成绩，成绩等级如下：60分及其以上为通过，60分以下为不通过

  #define _CRT_SECURE_NO_WARNINGS 1
  #include 
  int main()
  {
  	int d = 0;
  	scanf("%d", &d);
  	int n;
  	n = d > 60 ? 0 : 1;
  	switch (n)
  	{
  		case 0:
  			printf("合格\n");
  			break;
  		case 1:
  			printf("不合格\n");
  			break;
  	}
  	return 0;
  }

• 从键盘上输入学生成绩，成绩等级如下：
  90~100   “优”
  80~90   “良”
  70~80   “中”
  60~70   “合格”
  60以下   “不合格”

  #define _CRT_SECURE_NO_WARNINGS 1
  #include 
  int main()
  {
  	int d = 0;
  	scanf("%d", &d);
  	int n;
  	n = d > 60 ? 1: 0;
  	if (n)
  	{
  		d /= 10;
  		switch (d)
  		{
  		case 6:
  			printf("合格\n");
  			break;
  		case 7:
  			printf("中\n");
  			break;
  		case 8:
  			printf("良\n");
  			break;
  		case 9:
  			printf("优\n");
  			break;
  		case 10:
  			printf("优\n");
  			break;
  		}
  	}
  	else
  		printf("不合格");
  	return 0;
  }

while循环 

1. 求1到10之间奇数的和与偶数的和

   #define _CRT_SECURE_NO_WARNINGS 1
   #include 
   int main()
   {
   	int a = 0, b = 0, c = 1, d;
   	printf("请输入一个数字求 1~这个数字 范围内的偶数和奇数的和\n");
   	scanf("%d", &d);
   	while (c <= d)
   	{
   		if (c % 2 == 0)
   			a = a + c;
   		else
   			b = b + c;
   		c++;
   	}
   	printf("偶数和是 %d,奇数和是 %d\n", a, b);
   	return 0;
   }

2. ---

   求1/2+1/4+1/6+……的和，最后一项的值小于0.0001为止。

   #define _CRT_SECURE_NO_WARNINGS 1
   #include 
   int main()
   {
   	int a = 2;
   	double sum = 0;
   	while (a <= 10002)
   	{
   		sum = sum + 1.0 / a;
   		a += 2;
   	}
   	printf("%lf\n", sum);
   	return 0;
   }
   

3. 求和1/2-1/3+1/4-1/5+1/6……+1/100

   #define _CRT_SECURE_NO_WARNINGS 1
   #include 
   int main()
   {
       double b = 1.0, sum = 0, a;
       int i = 1;
      while(i <= 100)
       {
           a = b / i;
           sum += a;
           b = -b;
           i++;
       }
       printf("sum = %.2f\n", sum);
       return 0;
   }

4. 、求10！（求10的阶乘）

   #define _CRT_SECURE_NO_WARNINGS 1
   #include 
   int main()                
   {                            
       int p = 1;
       int n;
       int i = 2;
       printf("请输入所求阶乘数：\n");
       scanf("%d", &n);
       while(i <= n) 
       {
           p = p * i;
           ++i;
       }
       printf("%d! = %d", n, p);
       return  0;
   }