CCF 201503-3 Festival

subject

Problem description

　　 There is a kind of festival whose date is not fixed , But rather “a Month's Day b A few weeks c” In the form of , For example, mother's Day is set on the second Sunday of May every year .
　　 Now? , Here you are. a,b,c and y1, y2(1850 ≤ y1, y2 ≤ 2050), I hope you output from... Ad y1 Year to ad y2 Every year during the year a Month's Day b A few weeks c Date .
　　 Tips ： Rules about leap years ： The year is 400 Is a leap year , Otherwise the year is 4 Is not a multiple of 100 The multiple of is leap year , Other years are not leap years . for example 1900 A year is not a leap year , and 2000 Year is a leap year .
　　 For your convenience , It is known that 1850 year 1 month 1 It's Tuesday .

Input format

　　 The input contains exactly one line , There are five integers a, b, c, y1, y2. among c=1, 2, ……, 6, 7 Monday means Monday 、 Two 、……、 6、 ... and 、 Japan .

Output format

　　 about y1 and y2 Every year between , Include y1 and y2, Output a line in the order of the year from small to large .
　　 If this year's a Yue di b A few weeks c Do exist , with "yyyy/mm/dd" Format output , That is, the four digit year , Two digit months , Double digit date , Use a slash in the middle “/” Separate , If the number of digits is insufficient, fill in zero .
　　 If this year's a Yue di b A few weeks c There is no such thing as , The output "none"（ No double quotes ).

The sample input

5 2 7 2014 2015

Sample output

2014/05/11
2015/05/10

Evaluate use case size and conventions

　　 All evaluation cases meet ：1 ≤ a ≤ 12,1 ≤ b ≤ 5,1 ≤ c ≤ 7,1850 ≤ y1, y2 ≤ 2050.

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Problem analysis

This is about the date , First, consider each year in the given interval ： The idea is to get from 1850 year 1 month 1 How many days does it take to reach the target date TOTAL_DAYS, Then on 7 The remainder can be mapped to the day of the week （ This mapping is based on 1850 year 1 month 1 The day is set on Tuesday ）. In order to judge whether the date is legal , To put TOTAL_DAYS The specific date is , Here are some details . The following code is given directly .

Code

#include<cstdio>
using namespace std;

int AbnormalYear(int year){
	if((year%4==0&&year%100!=0)||year%400==0)
		return 1;
	return 0;
}

bool Legal(int abnormal,int mon,int day){
	switch(mon){
		case 4:
			if(day<=30)
				return true;
			break;
		case 6:
			if(day<=30)
				return true;
			break;
		case 9:
			if(day<=30)
				return true;
			break;
		case 11:
			if(day<=30)
				return true;
			break;
		case 2:
			if(day<=28+abnormal)
				return true;
			break;
		default:
			if(day<=31)
				return true;
	}
	return false;
}

int main(){
	int month_total[13];
	month_total[1] = 0;
	month_total[2] = 31;
	month_total[3] = 59;
	month_total[4] = 90;
	month_total[5] = 120;
	month_total[6] = 151;
	month_total[7] = 181;
	month_total[8] = 212;
	month_total[9] = 243;
	month_total[10] = 273;
	month_total[11] = 304;
	month_total[12] = 334;
	
	int abnormal_month_total[13];
	abnormal_month_total[1] = 0;
	abnormal_month_total[2] = 31;
	abnormal_month_total[3] = 60;
	abnormal_month_total[4] = 91;
	abnormal_month_total[5] = 121;
	abnormal_month_total[6] = 152;
	abnormal_month_total[7] = 182;
	abnormal_month_total[8] = 213;
	abnormal_month_total[9] = 244;
	abnormal_month_total[10] = 274;
	abnormal_month_total[11] = 305;
	abnormal_month_total[12] = 335;
	
	int mon,n,weekday,from,to,sum = 0,t1,t2,t3;
	const int week[7] = {2,3,4,5,6,7,1};
	
	scanf("%d %d %d %d %d",&mon,&n,&weekday,&from,&to);
	
	sum += (from - 1850) * 365;
	for(int i=1850;i<from;i++)
		sum += AbnormalYear(i);
	
	for(int i=from;i<=to;i++){
		if(!AbnormalYear(i)){
			t1 = sum + month_total[mon];// First day of the month  
			t2 = week[t1%7];// That month 1 What day is the th  
			t3 = 1 + (n - 1) * 7;
			if(weekday<t2){
				t3 += 7 - t2 + weekday;// The day of the month  
			}
			else if(weekday>t2){
				t3 += weekday - t2;
			}
			//else t3 = t3; 
			if(Legal(0,mon,t3))
				printf("%d/%02d/%02d\n",i,mon,t3);
			else
				printf("none\n");
			sum += 365;
		}
		else{
			t1 = sum + abnormal_month_total[mon];// First day of the month  
			t2 = week[t1%7];// That month 1 What day is the th  
			t3 = 1 + (n - 1) * 7;
			if(weekday<t2){
				t3 += 7 - t2 + weekday;// The day of the month  
			}
			else if(weekday>t2){
				t3 += weekday - t2;
			}
			//else t3 = t3; 
			if(Legal(1,mon,t3))
				printf("%d/%02d/%02d\n",i,mon,t3);
			else
				printf("none\n");
			sum += 366;
		}
	}
	
	return 0;
}