C语言实现日期转换小工具

已知日期转天数；已知天数转日期
核心是判断闰年，可见：C语言判断闰年

//编译环境vs 2019
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <iostream>
using namespace std;
#include <string>

int year = 0, sum = 0, month = 0, day = 0;

bool leapYear(int year);
int leapMonth[12] = {
    
     31,29,31,30,31,30,31,31,30,31,30,31 };
int noleapMonth[12] = {
    
     31,28,31,30,31,30,31,31,30,31,30,31 };
void schema1();
void schema2();

void main()
{
    
    
    int year = 0, sum = 0, month = 0, day = 0;
    int key = 0;
    printf("1.日期转天数：输入x年x月x日，输出第x天\n2.天数转日期：输入x年第x天，输出x年x月x日\n");
    scanf("%d",&key);
    switch (key)
    {
    
    
    case 1:schema1(); break;
    case 2:schema2(); break;
    default:break;
    }
}

void schema1()
{
    
    
    printf("\n请输入：xxxx-xx-xx：\n");
    string str;
    cin >> str;

    int year, month, day;
    year = stoi(str.substr(0, 4));
    month = stoi(str.substr(5, 7));
    day = stoi(str.substr(8, 10));
    //printf("%d %d %d", year, month, day);
    //至此，年，月，日已经安排好

    int sum = 0;
    if (leapYear(year))   //如果是闰年，则一年有366天
    {
    
    //先判断是否是闰年，再判断是哪一个月
        for (int k = 0; k < month - 1; k++)
        {
    
    
            sum += leapMonth[k];
        }
        sum += day;
    }
    else
    {
    
    
        for (int k = 0; k < month - 1; k++)
        {
    
    
            sum += noleapMonth[k];
        }
        sum += day;
    }

    printf("\n%d\n", sum);
}

void schema2()
{
    
    
    printf("\n请输入：xxxx xxx：\n");
    scanf("%d", &year);
    scanf("%d", &sum);
    if (leapYear(year))       //先判断是否闰年
    {
    
    
        // printf("闰年\n");
        for (int i = 0; sum > 0; i++)
        {
    
    
            month += 1;

            day = sum;
            sum -= leapMonth[i];
        }
    }

    else
    {
    
    
        // printf("闰年\n");
        for (int i = 0; sum > 0; i++)
        {
    
    
            month += 1;
            day = sum;
            sum -= noleapMonth[i];

        }
    }
    printf("\n%d-%d-%d\n", year, month, day);

}

bool leapYear(int year)
{
    
    
    if(year % 4 == 0 && year % 100 != 0|| year % 400 == 0)
        return true;
    else
        return false;
}