解析C++编程中的选择结构和switch语句的用法

#include <iostream>
using namespace std;
int main( )
{
  int year;
  bool leap;
  cout<<"please enter year:";//输出提示
  cin>>year; //输入年份
  if (year%4==0) //年份能被4整除
  {
   if(year%100==0)//年份能被4整除又能被100整除
   {
     if (year%400==0)//年份能被4整除又能被400整除
      leap=true;//闰年,令leap=true(真)
     else
      leap=false;
   } //非闰年,令leap=false(假)
   else //年份能被4整除但不能被100整除肯定是闰年
     leap=true;
  } //是闰年,令leap=true
  else //年份不能被4整除肯定不是闰年
   leap=false; //若为非闰年,令leap=false
  if (leap)
   cout<<year<<" is "; //若leap为真,就输出年份和“是”
  else
   cout<<year<<" is not ";//若leap为真,就输出年份和“不是”
  cout<<" a leap year."<<endl; //输出“闰年”
  return 0;
}

① 2005↙
2005 is not a leap year.
② 1900↙
1900 is npt a leap year.

if(year%4!=0)
  leap=false;
else if(year%100!=0)
  leap=true;
else if(year%400!=0)
  leap=false;
else
  leap=true;

  s<250km 没有折扣
  250≤s<500  2%折扣
  500≤s<1000  5%折扣
  1000≤s<2000  8%折扣
  2000≤s<3000  10%折扣
  3000≤s  15%折扣

  f = p * w * s * (1 - d)

#include <iostream>
using namespace std;
int main( )
{
  int c,s;
  float p,w,d,f;
  cout<<"please enter p,w,s:";
  cin>>p>>w>>s;
  if(s>=3000)
   c=12;
  else
   c=s/250;
  switch (c)
  {
   case 0:d=0;break;
   case 1:d=2;break;
   case 2:
   case 3:d=5;break;
   case 4:
   case 5:
   case 6:
   case 7:d=8;break;
   case 8:
   case 9:
   case 10:
   case 11:d=10;break;
   case 12:d=15;break;
  }
  f=p*w*s*(1-d/100.0);
  cout<<"freight="<<f<<endl;
  return 0;
}

please enter p,w,s:100 20 300↙
freight=588000

switch(表达式)
{
  case 常量表达式1:语句1
  case 常量表达式2:语句2
  ...
  case 常量表达式n:语句n
  default:语句n+1
 }

switch(grade)
{
  case 'A': cout<<"85~100\n";
  case 'B': cout<<"70~84\n";
  case 'C': cout<<"60~69\n";
  case 'D': cout<<"<60\n";
  default: cout<<"error\n";
}

  85~100
  70~84
  60~69
  <60
  error

switch(grade)

{
  case 'A': cout<<"85~100\n";break;
  case 'B': cout<<"70~84\n";break;
  case 'C': cout<<"60~69\n";break;
  case 'D': cout<<"<60\n";break;
  default: cout<<"error\n";break;
}

  case 'A':
  case 'B':
  case 'C': cout<<">60\n";break;
  ...