纪念日问题

#include<stdio.h>
#include<string.h>
#include<math.h>

bool IsLeapYear(int year);
int main()
{
    
    
	long minute = 0;
	long day = 0;
	int i = 0;
	for(i = 1922; i<=2019; i++){
    
    
		if(IsLeapYear(i)){
    
    
			day+=366;
		}
		else{
    
    
			day+=365;
		}
	}
	minute = (day+344)*24*60;
	printf("%ld",minute);
	return 0;
}

bool IsLeapYear(int year){
    
    
	return (year%400==0||(year%4==0&&(year%100!=0)));
}

//1 3 5 7 8 10 12 (31)
//4 6 9 11 (30)
//365-(31+28+31+30+31+30+22)=? 1921剩余 
//31+29+31+30+31+30= ? 2020度过