用综合的方法判断为概念的第几天（1）

/**Copyright (c) 2013 ,烟台大学计算机学院*All rights reserved.*作者：王至超*完成日期：2013年11月12*版本号：v1.0*问题描述：输入年月日。判断为该年的第几天*样例输入：2000 3 1*样例输出：为第61天*问题分析：用简单的方法，学会活学活用*/#include <iostream>using namespace std;int main(){int year,n;   cout<<"请输入任意一个年份year：";      cin>>year;       if (year%400==0)      {            n=29;           cout<<n<< endl; }       else if (year%4==0&&year%100!=0)          {  n=29;               cout <<n<<endl;           }           else           {  n=28;               cout<<n<<endl;          }   int month,m,sum;   cin>>month;   switch (month){   case 1:    m=0;    cout<<m<<endl;break;    case 2:    m=31;    cout<<m<<endl;break;    case 3:    m=31;    cout<<m<<endl;break;    case 4:    m=31*2;    cout<<m<<endl;break;    case 5:    m=31*2+30;    cout<<m<<endl;break;    case 6:    m=31*3+30;    cout<<m<<endl;break;    case 7:    m=31*3+30*2;    cout<<m<<endl;break;    case 8:    m=31*4+30*2;    cout<<m<<endl;break;    case 9:    m=31*5+30*2;    cout<<m<<endl;break;    case 10:    m=31*5+30*3;    cout<<m<<endl;break;    case 11:    m=31*6+30*3;    cout<<m<<endl;break;    case 12:    m=31*6+30*4;    cout<<m<<endl;break;}int day;cin>>day;   if(month<=2)

{sum=m+day;}

else

{sum=m+day+n;}

     cout<<"为该年第"<<sum<<"天"<<endl;return 0;}