2014 第三章上机实验2 switch语句习题 答案(2)

C++程序设计(第2版)

break;

case '-':cout<

case '*':cout<

cout<<\除数不能为0\ else

cout<

default: cout<<\运算符错误！\ } return 0; }

解2：

#include int main() { cout<<\输入操作数1，运算符，操作数2\ int a,b,d; char op; cin>>a>>op>>b; switch(op) { case '+':d=a+b;break; case '-':d=a-b;break; case '*':d=a*b;break; case '/': if(b==0) { cout<<\除数不能为0\ return 0; } else d=a/b; break; } cout<

C++程序设计(第2版)

解3：

#include int main() { cout<<\输入操作数1，运算符，操作数2\ int a,b,d; char op; cin>>a>>op>>b; if((op=='/')&&(b==0)) cout<<\除数不能为零\ else { switch(op) { case '+':d=a+b;break; case '-':d=a-b;break; case '*':d=a*b;break; case '/':d=a/b;break; } cout<

4、设计一个程序，根据用户输入的年、月，打印出该年、该月的天数。 解1

#include int main()

{ int year,month,days;

cout<<\ cin>>year>>month; switch (month) {

case 1: case 3: case 5: case 7: case 8: case 10:

case 12:days=31;

C++程序设计(第2版)

break; case 4: case 6: case 9:

case 11:days=30; break;

case 2:if (year @0==0|| year %4==0 && year 0!=0) days=29; else

days=28; break; }

cout<<\} 解2

#include void main() { int x,y,z,w; cout<<\请输入年月：\ cin>>x>>y; if(x%4==0&&x0!=0||x@0==0) { z=366; switch(y) { case 1:; case 3:; case 5:; case 7:; case 8:; case 10:; case 12:w=31;break; case 2:w=29;break; case 6:; case 4:; case 9:; case 11:w=30;break;

C++程序设计(第2版)

} } else { z=365; switch(y) { case 1:; case 3:; case 5:; case 7:; case 8:; case 10:; case 12:w=31;break; case 2:w=28;break; case 6:; case 4:; case 9:; case 11:w=30;break; } } cout<<\ cout<<\本年的天数为：\本月的天数为：\} 解3

#include int main()

{ int year,month,days,ydays; bool isleap;

cout<<\ cin>>year>>month; isleap=(year @0==0|| year %4==0 && year 0!=0); if (isleap) ydays=366; else ydays=365; switch (month) {

case 1: case 3:

C++程序设计(第2版)

case 5: case 7: case 8: case 10:

case 12:days=31; break; case 4: case 6: case 9:

case 11:days=30; break;

case 2:if (isleap) days=29; else

days=28; break; }

cout<

5. 企业发放的奖金来自利润提成。利润i低于或等于10万元的,可提成10%为奖金;利润i高于10万元,低于或等于20万元(100000

#include using namespace std; int main () {long i;

float bonus,bon1,bon2,bon4,bon6,bon10; int c;

bon1=100000*0.1;

bon2=bon1+100000*0.075; bon4=bon2+200000*0.05; bon6=bon4+200000*0.03; bon10=bon6+400000*0.015;

C++程序设计(第2版)

cout<<\ cin>>i; c=i/100000; if (c>10) c=10; switch(c)

{case 0: bonus=i*0.1; break;

case 1: bonus=bon1+(i-100000)*0.075; break; case 2:

case 3: bonus=bon2+(i-200000)*0.05;break; case 4:

case 5: bonus=bon4+(i-400000)*0.03;break; case 6: case 7: case 8:

case 9: bonus=bon6+(i-600000)*0.015; break; case 10: bonus=bon10+(i-1000000)*0.01; }

cout<<\ return 0; }