*6-2 CCF 2015-03-3 Festival

Title Description

Source code

#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;

const int N = 1000;
int monthdays[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} };
int isleapyear(int year)
{
    
	return ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0)) ? 1 : 0;
}
int main()
{
    
	int a, b, c, y1, y2;
	int weekd, d, leap;
	cin >> a >> b >> c >> y1 >> y2;
	int basedays = 0;
	// Calculate from 1850 Year to y1 The days of the year 
	//for(int i=1850;i<y1;i++)
		//basedays+=365+isleapyear(i);
	basedays = (y1 - 1850) * 365 + (y1 - 1850) / 4 - (y1 - 1850) / 100 + (y1 - 1850) / 400;
	// Calculate the month, year and date and output 
	for (int i = y1; i <= y2; i++)
	{
    
		int base_weekday = 2;// Every time a new year is calculated   It's all about 1850 year 1 month 1 Day as base  ： Tuesday 
		leap = isleapyear(i);
		// Calculation i year a month 1 Day is the day of the year 
		int days = 0;
		for (int j = 1; j < a; j++)
			days += monthdays[leap][j];
		// hold 1850 year 1 month 1 Japan   To i year a month 1 The total number of days per day is the same as 7 Seeking remainder 
		base_weekday += (basedays + days) % 7;
		if (base_weekday > 7)
			base_weekday %= 7;//i year a month 1 What day is the day of the week 
		int d = 1, num = 0;//d Record the number of days in the month  num  Record the week of the month c
		while (1)
		{
    
			if (base_weekday == c)
			{
    
				num++;
			}
			if (num == b) break;
			base_weekday++;
			if (base_weekday > 7) base_weekday = 1;
			d++;
		}
		if (d > monthdays[leap][a])
			cout << "none" << endl;
		else
		{
    
			cout << i << "/";
			if (a < 10) cout << "0";
			cout << a << "/";
			if (d < 10) cout << "0";
			cout << d << endl;
		}
		basedays += 365 + leap;
	}
	return 0;
}