一、实验目的

1．学会正确使用逻辑运算符和逻辑表达式；
2．熟练掌握if语句和switch 语句。

二、实验内容

1．任意输入4个整数，从大到小排列输出。
2．输入一组成绩，要求输出成绩等级为A，B，C，D，E。其中90—100为A，80—89为B，70—79为C，60—69为D，60以下为E。
3．有一个函数

写一程序，输出y值。

三、实验记录

3.1 四个数字排序

（1）源代码

# include <stdio.h>

int main(void)
{
	int a,b,c,d,t;
	printf("Please enter the data(a,b,c,d):\n");
	scanf("%d %d %d %d",&a,&b,&c,&d);

	if(a<b)
	{
		t = a;
		a = b;
		b = t;
	}
	if(a<c)
	{
		t = a;
		a = c;
		c = t;
	}
	if(a<d)
	{
		t = a;
		a = d;
		d = t;
	}	
if(b<c)
	{
		t = b;
		b = c;
		c = t;
	}
	if(b<d)
	{
		t = b;
		b = d;
		d = t;
	}
	if(c<d)
	{
		t = c;
		c = d;
		d = t;
	}

	printf("The retult is: %d %d %d %d\n",a,b,c,d);
	return 0;
}

（2）运行结果截图

3.2 判断分数等级

（一）源代码

# include <stdio.h>
/*函数声明*/
void inputgrade(int * grade,int lg);
int main(void)
{
	int grade[10];
	int lg,i;
	printf("How many of the data?\n");
	scanf("%d",&lg);
	printf("Please enter the grade:\n");
	inputgrade(grade,lg);
	for(i = 0;i<lg;++i)
	{
		if(grade[i]<=100 && grade[i]>=90)
			printf("The %dth degree is A!\n",i+1);
		else if(grade[i]<90 && grade[i]>=80)
			printf("The %dth degree is B!\n",i+1);
		else if(grade[i]<80 && grade[i]>=70)
			printf("The %dth degree is C!\n",i+1);
		else if(grade[i]<70 && grade[i]>=60)
			printf("The %dth degree is D!\n",i+1);
		else if(grade[i]<60)
			printf("The %dth degree is E!\n",i+1);
		else
			printf("The %dth data is wrong!");
	}
	return 0;
}
/*输入数据的函数，当用户输入0时表示输入已完成。*/
void inputgrade(int * grade,int lg)
{
	int i;
	for(i = 0;i<lg;++i)
	{
		printf("Please enter the %dth data:\n",i+1);
		scanf("%d",grade+i);
	}
	return;
}

（2）运行结果截图

3.3 已知x的值求y的值

（1）程序框图

（2）源代码

# include <stdio.h>

int main(void)
{
	int x,y;
	printf("Please enter x:\n");
	scanf("%d",&x);
	if(x<1)
		printf("y = %d\n",x);
	else if(x>=1 && x<10)
		printf("y = %d\n",2*x+1);
	else
		printf("y = %d\n",3*x-1);
	return 0;
}

（3）运行结果截图



3.4 设计程序计算需要如何付钱
（1）源代码

# include <stdio.h>
int main(void)
{
	int a = 0,b = 0,c = 0,d = 0,e = 0,f = 0,g = 0,h = 0,k = 0,i,j;
	float mon;
	printf("How much the money we need?\n");
	scanf("%f",&mon);

	/*判断需要多少张100元的钞票*/
	while(mon>=100)
	{
		a++;
		mon = mon - 100;
	}
	/*判断需要多少张50元的钞票*/
	while(mon>=50)
	{
		b++;
		mon = mon - 50;
	}
	/*判断需要多少张10元的钞票*/
	while(mon>=10)
	{
		c++;
		mon = mon - 10;
	}
	/*判断需要多少张5元的钞票*/
	while(mon>=5)
	{
		d++;
		mon = mon - 5;
	}
	/*判断需要多少张2元的钞票*/
	while(mon>=2)
	{
		e++;
		mon = mon - 2;
	}
	/*判断需要多少张1元的钞票*/
	while(mon>=1)
	{
		f++;
		mon = mon - 1;
	}

	/*判断需要多少张0.1元的钞票*/
	while(mon>=0.1)
	{
		g++;
		mon = mon - 0.1;
	}
/*判断需要多少张0.05元的钞票*/
	while(mon>=0.05)
	{
		h++;
		mon = mon - 0.05;
	}
	
	/*判断需要多少张0.01元的钞票*/
	while(mon>0)
	{
		k++;
		mon = mon - 0.01;
	}
	printf("We need:\n");
	printf("100 yuan %d page(s)\n50 yuan %d page(s)\n10 yuan %d page(s)\n5 yuan %d page(s)\n2 yuan %d page(s)\n1 yuan %d page(s)\n0.1 yuan %d page(s)\n0.05 yuan %d page(s)\n0.01 yuan %d page(s)\n",a,b,c,d,e,f,g,h,k);

	return 0;
}

（2）程序运行截图