按你的思路大概是这样:
int year = 0, min = 0, day = 365, n = 0, sum = 0;
for(year = 1921; year <= 2020; year++)
{
// 能整除400或者整除4且不能整除100的年份为闰年
if(((year % 400) == 0) || (((year % 4) == 0) && ((year % 100) != 0)))
{
n++;
}
}
sum = (2020 - 1921) * day + n - 22;
min = sum * 24 * 60;
printf("%d\n", min);
仅供参考