第三章 3.4
main()
{int a,b,c
long int u,n
float x,y,z
char c1,c2
a=3b=4c=5
x=1.2y=2.4z=-3.6
u=51274n=128765
c1='a'c2='b'
printf("\n")
printf("a=%2d b=%2d c=%2d\n",a,b,c)
printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z)
printf("x+y=%5.2f y+z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x)
printf("u=%6ld n=%9ld\n",u,n)
printf("c1='%c'or %d(ASCII)\n",c1,c1)
printf("c2='%c'or %d(ASCII)\n",c2,c2)
}
3.5
57
5 7
67.856400,-789.123962
67.856400,-789.123962
67.86 -789.12,67.856400,-789.123962,67.856400,-789.123962
6.785640e+001,-7.89e+002
A,65,101,41
1234567,4553207,d687
65535,177777,ffff,-1
COMPUTER, COM
3.6
a=3 b=7/
x=8.5 y=71.82/
c1=A c2=a/
3.7
10 20Aa1.5 -3.75 +1.4,67.8/
(空 3)10(空3)20Aa1.5(空1)-3.75(空1)(随意输入一个数），67.8 回车
3.8
main()
{float pi,h,r,l,s,sq,sv,sz
pi=3.1415926
printf("input r,h\n")
scanf("%f,%f",&r,&h)
l=2*pi*r
s=r*r*pisq=4*pi*r*r
sv=4.0/3.0*pi*r*r*r
sz=pi*r*r*h
printf("l=%6.2f\n",l)
printf("s=%6.2f\n",s)
printf("sq=%6.2f\n",sq)
printf("vq=%6.2f\n",sv)
printf("vz=%6.2f\n",sz)
}
3.9
main()
{float c,f
scanf("%f",&f)
c=(5.0/9.0)*(f-32)
printf("c=%5.2f\n",c)
}
3.10
#include"stdio.h"
main()
{char c1,c2
scanf("%c,%c",&c1,&c2)
putchar(c1)
putchar(c2)
printf("\n")
printf("%c%c\n",c1,c2)
}
4.3
(1)0 (2)1 (3)1 (4)0 (5)1
4.4
main()
{int a,b,c
scanf("%d,%d,%d",&a,&b,&c)
if(a<b)
if(b<c)
printf("max=%d\n",c)
else
printf("max=%d\n",b)
else if(a<c)
printf("max=%d\n",c)
else
printf("max=%d\n",a)
}
main() {int a,b,c,temp,max
scanf("%d,%d,%d",&a,&b,&c)
temp=(a>b)?a:b
max=(c>temp)?c:temp
printf("max=%d",max)
}
4.5
main()
{int x,y
scanf("%d",&x)
if(x<1)y=x
else if(x<10)y=2*x-1
else y=3*x-11
printf("y=%d",y)
}
4.6
main()
{int score,temp,logic
char grade
logic=1
while(logic)
{scanf("%d",&score)
if(score>=0&&score<=100)logic=0
}
if(score==100)
temp=9
else
temp=(score-score%10)/10
switch(temp)
{case 9:grade='A'break
case 8:grade='B'break
case 7:grade='C'break
case 6:grade='D'break
case 5:
case 4:
case 3:
case 2:
case 1:
case 0:grade='E'
}
printf"score=%d,grade=%c",score,grade)
}
4.7 main()
{long int num
int indiv,ten,hundred,thousand,ten_thousand,place
scanf("%ld",&num)
if(num>9999) place=5
else if(num>999) place=4
else if(num>99) place=3
else if(num>9) place=2
else place=1
printf("place=%d\n",place)
ten_thousand=num/10000
thousand=(num-ten_thousand*10000)/1000
hundred=(num-ten_thousand*10000-thousand*1000)/100
ten=(num-ten_thousand*10000-thousand*1000-hundred*100)/10
indiv=num-ten_thousand*10000-thousand*1000-hundred*100-ten*10
switch(place)
{case 5:printf("%d,%d,%d,%d,%d\n",ten_thousand,thousand,hundred,ten,indiv)
printf("%d,%d,%d,%d,%d\n",indiv,ten,hundred,thousand,ten_thousand)
break
case 4:printf("%d,%d,%d,%d\n",thousand,hundred,ten,indiv)
printf("%d,%d,%d,%d\n",indiv,ten,hundred,thousand)
break
case 3:printf("%d,%d,%d\n",hundred,ten,indiv)
printf("%d,%d,%d\n",indiv,ten,hundred)
break
case 2:printf("%d,%d\n",ten,indiv)
printf("%d,%d\n",indiv,ten)
break
case 1:printf("%d\n",indiv)
printf("%d\n",indiv)
}
}
4.8
main()
{long i
float bonus,bon1,bon2,bon4,bon6,bon10
bon1=100000*0.1
bon2=bon1+100000*0.075
bon4=bon2+200000*0.05
bon6=bon4+200000*0.03
bon10=bon6+400000*0.015
scanf("%ld",&i)
if(i<=1e5)bonus=i*0.1
else if(i<=2e5)bonus=bon1+(i-100000)*0.075else if(i<=4e5)bonus=bon2+(i-200000)*0.05
else if(i<=6e5)bonus=bon4+(i-400000)*0.03
else if(i<=1e6)bonus=bon6+(i-600000)*0.015
else bonus=bon10+(i-1000000)*0.01
printf("bonus=%10.2f",bonus)
}
main()
{long i
float bonus,bon1,bon2,bon4,bon6,bon10
int branch
bon1=100000*0.1
bon2=bon1+100000*0.075
bon4=bon2+200000*0.05
bon6=bon4+200000*0.03
bon10=bon6+400000*0.015
scanf("%ld",&i)
branch=i/100000
if(branch>10)branch=10
switch(branch)
{case 0:bonus=i*0.1break
case 1:bonus=bon1+(i-100000)*0.075break
case 2:
case 3:bonus=bon2+(i-200000)*0.05break
case 4:
case 5:bonus=bon4+(i-400000)*0.03break
case 6:
case 7
case 8:
case 9:bonus=bon6+(i-600000)*0.015break
case 10:bonus=bon10+(i-1000000)*0.01
}
printf("bonus=%10.2f",bonus)
}
4.9
main()
{int t,a,b,c,d
scanf("%d,%d,%d,%d",&a,&b,&c,&d)
if(a>b){t=aa=bb=t}
if(a>c){t=aa=cc=t}
if(a>d){t=aa=dd=t}
if(b>c){t=bb=cc=t}
if(b>d){t=bb=dd=t}
if(c>d){t=cc=dd=t} printf("%d %d %d %d\n",a,b,c,d)
}
4.10
main()
{int h=10
float x,y,x0=2,y0=2,d1,d2,d3,d4
scanf("%f,%f",&x,&y)
d1=(x-x0)*(x-x0)+(y-y0)*(y-y0)
d2=(x-x0)*(x-x0)+(y+y0)*(y+y0)
d3=(x+x0)*(x+x0)+(y-y0)*(y-y0)
d4=(x+x0)*(x+x0)+(y+y0)*(y+y0)
if(d1>1&&d2>1&&d3>1&&d4>1)h=0
printf("h=%d",h)
}
while (--lim >0 &&(c=getchar()) != EOF &&c!= '\n')这一句的 --lim >0 // 这个是最大长度呀。这里是#define MAXLINE 1000 /* 定义接受输入的最大行数*/
--lim >0 // 对读取的总长度进行限制，以免访问数组越界。
(c=getchar()) != EOF // 是否读取到文件末尾，文件读完毕，即使没有读到最大长度也结束了。
c!= '\n') // 是否是换行，读取到换行也意味着结束。
这是《C Primer Plus》第四章编程练习（第六版）4.8.5的练习题：
参考代码如下：
#include <stdio.h>int main(void){
float downloadSpeed,fileSize
printf("Please input your download speed(Mb/s):")
scanf("%f",&downloadSpeed)
printf("Please input you file size(Mb):")
scanf("%f",&fileSize)
printf("At %.2f megabits per second, a file of %.2f megabytes\n",downloadSpeed,fileSize)
printf("download in %.2f seconds.\n",fileSize*8/downloadSpeed)
}
执行结果如下：
