Zoj 2420 calendar stores the struct in the vector

Question address: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemid = 1420

Ideas:

At the beginning, I thought the question was simple, but different months had different days, and I don't know how to solve it easily. Then I thought of the table-making method, and used the information of each day as a struct, which exists in the vector one by one. The angle mark of the vector is auto-incrementing, that is, the number of days in.

Note:

1 leap year judgment

2 The output format 1-1 must be written as 01-01

3. for Super-memory problems, set the three attribute values to short int, and the direct int value will exceed the memory.

4 weeks, because 2000-01-01 is Saturday, the nth day is week (n + 6) % 7.

Code:

#include<iostream>#include<string>#include<vector>using namespace std;struct date{  short year;  short month;  short day;};bool isleap(int n){ if(n%4==0&&n%100!=0)  return 1; if(n%400==0)   return 1; return 0;}int shortmonth(int n){   if(n==2||n==4||n==6||n==9||n==11)   return 1;   else   return 0;}int main(){  int n;  string * day=new string[7];  day[0]="Monday";  day[1]="Tuesday";  day[2]="Wednesday";  day[3]="Thursday";  day[4]="Friday";  day[5]="Saturday";  day[6]="Sunday";  vector<date> v;   for(int i=2000;i<10000;i++)       for(int j=1;j<=12;j++)         for(int k=1;k<=31;k++)      {           if(j==2&&isleap(i)&&k>29) continue;           if(j==2&&(!isleap(i))&&k>28) continue;           if(shortmonth(j)&&k>30)  continue;           date today;           today.year=i;           today.month=j;           today.day=k;           v.push_back(today);        }  while(cin>>n)  {     if(n==-1)  break;     cout<<v[n].year<<"-";     if(v[n].month<10)     cout<<"0";     cout<<v[n].month<<"-";     if(v[n].day<10)     cout<<"0";     cout<<v[n].day<<" ";     cout<<day[(n+5)%7]<<endl;  }}