CCF 201509-2 date calculation

subject

Problem description

　　 Given a year y And an integer d, Ask the first of the year d What's the day ？
　　 Pay attention to the leap year 2 Monthly 29 God . One of the following conditions is leap year ：
　　1） The year is 4 Integer multiple , And it's not 100 Integer multiple ;
　　2） The year is 400 Integer multiple .

Input format

　　 The first line of input contains an integer y, Indicates the year , The year is 1900 To 2015 Between （ contain 1900 and 2015）.
　　 The second line of input contains an integer d,d stay 1 to 365 Between .

Output format

　　 Output two lines , One integer per row , The month and date of the answer respectively .

The sample input

2015
80

Sample output

3
21

The sample input

2000
40

Sample output

2
9

---

Problem analysis

A basic date type topic , Go straight to the code .

Code

#include<cstdio>
using namespace std;

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

int main(){
	int year,day,i,t;
	int month[2][13];//[1][x]- Leap year  [0][x]- Non leap year  
	month[0][1] = month[1][1] = 0;
	month[0][2] = month[1][2] = 31;
	month[0][3] = 59;
	month[0][4] = 90;
	month[0][5] = 120;
	month[0][6] = 151;
	month[0][7] = 181;
	month[0][8] = 212;
	month[0][9] = 243;
	month[0][10] = 273;
	month[0][11] = 304;
	month[0][12] = 334;
	for(int i=3;i<=12;i++)
		month[1][i] = month[0][i] + 1;
	
	scanf("%d %d",&year,&day);
	t = AbnormalYear(year);
	for(i=12;month[t][i] >= day;i--);
	printf("%d\n%d\n",i,day-month[t][i]);
	return 0;
}