C 解法, 执行用时: 1ms, 内存消耗: 340KB, 提交时间: 2021-09-12

#include <stdio.h>

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

int main()
{
    int daynum_of_leapyear[12] = {31,29,31,30,31,30,31,31,30,31,30,31};
    int daynum_of_commonyear[12] = {31,28,31,30,31,30,31,31,30,31,30,31};
    
    int year,month,day;
    scanf("%d", &year);
    scanf("%d", &month);
    scanf("%d", &day);
    
    int daynum = 0;
    if(1 == isleapyear(year))
    {
        for(int i =0; i<month-1; i++)
        {
            daynum += daynum_of_leapyear[i];
        }
    }
    else
    {
        for(int i =0; i<month-1; i++)
        {
            daynum += daynum_of_commonyear[i];
        }
    }
    
    daynum += day;
    printf("%d",daynum);
    return 0;
}

C 解法, 执行用时: 1ms, 内存消耗: 348KB, 提交时间: 2021-09-09

#include<stdio.h>
int main(){
    int year, month, day;
    int md[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    int res = 0;
    scanf("%d %d %d", &year, &month, &day);
    //check if run year
    if(year %4 == 0 && year % 100 != 0 || year % 40 == 0){
        md[1] = 29;
    }
    for(int i = 0; i < month - 1; i++){
        res = res + md[i];
    }
    res += day;
    printf("%d", res);
    return 0;
}