打渔晒网

//当天的前一年离1990年有几天，再加上当年的天数，余数 1 2 3则打鱼，否则晒网
#include <iostream>
#include <cstring>
using namespace std;

const int a[2][13] = {0,31,28,31,30,31,30,31,31,30,31,30,31,
0,31,29,31,30,31,30,31,31,30,31,30,31} ;

typedef struct Node
{
     int year,month,day;
}Node;

int is_leap(int year)
{
     if(year%4==0&&year%100!=0||year%400 == 0)//符合短路表达式规则
          return 1;
     return 0;
} 

int sum_days(Node today)
{
     int sum = 0;
     int i,j,k;
     for(i = 1990; i<today.year; i++)//没加等号，当天的前一年离1990年有几天，再加上当年的天数 
     {
          int flag = is_leap(i);
          if(flag)
               sum += 366;
          else
               sum += 365;
     }
     int tag = is_leap(today.year);
     for(i=1; i<today.month; i++)
          sum += a[tag][i];
     sum += today.day;//加上当月的天数 ,即便输入1990 1 1日， 则sum值为1,所以余数为1 2 3时打鱼，不是 0 1 2 
     return sum;
}

int main()
{
     int i,j,k;
     Node today;
     cout<<"Input date:"<<endl;
     while(cin>>today.year>>today.month>>today.day)
     {
          int sum = sum_days(today);
          int mod = sum%5;
          if(mod>0&&mod<4)
               cout<<"打鱼"<<endl;
          else
               cout<<"晒网"<<endl;
     }
     return 0;
}