poj 3393 Lucky and Good Months by Gregorian Calendar(模拟)

题目：http://poj.org/problem?id=3393
一道题目挺长的模拟题，参考了网上大神的题解。

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 using namespace std;

 int Month[]={,,,,,,,,,,,,};
 int Lmonth[]={,,,,,,,,,,,,};

 int check(int year)
 {
     if(year<)
     {
         if(year%==)
         return ;
         else
         return ;
     }
     else
     {
         if(year%==||(year%==&&year%!=)||year==)
         return ;
         else
         return ;
     }
 }

 int computeday(int year,int month)
 {
     int day=,i,j;
     for(i=; i<year; i++)
     if(check(i))
     day+=;
     else
     day+=;

     for(j=; j<month; j++)
     if(check(i))
     day+=Lmonth[j];
     else
     day+=Month[j];

     if(year>||(year==&&month>))
     day-=;
     return day;
 }

 int main()
 {
     int luck,good,t,*x;
     int ys,ms,ye,me;
     int day,i,j;
     cin>>t;
     while(t--)
     {
         luck=; good=;
         cin>>ys>>ms>>ye>>me;
         day=computeday(ys,ms);
         if((day+)%<=||(day+)%==)
         good++;

         for(i=ys; i<ye; i++)
         {
             if(check(i))
             x=Lmonth;
             else
             x=Month;
             if(i==ys)
             j=ms;
             else
             j=;
             for(; j<=; j++)
             {
                day+=*(x+j);
                if(i==&&j==)
                day-=;
                if((day+)%<=||(day+)%==)
               {
                good++;
                luck++;
               }
             }
         }

         if(check(i))
             x=Lmonth;
             else
             x=Month;
             if(i==ys)
             j=ms;
             else
             j=;
             for(; j<=me; j++)
             {
                day+=*(x+j);
                if(i==&&j==)
                day-=;
                if((day+)%<=||(day+)%==)
                {
                 if(j!=me)
                 good++;
                luck++;
                }
             }
             printf("%d %d\n",luck,good);
     }
     return ;
 }