简单易懂的代码

#include <bits/stdc++.h>
using namespace std;
int a,b,c,d,e,f,sum = 0;
int main(){
	cin >> a >> b >> c >> d >> e >> f;
	while(!(a == d && b == e && c == f)){
		c++;
		sum++;
		if((b == 1 || b == 3 || b == 5 || b == 7 || b == 8 || b == 10 || b == 12) && c == 32){
			c = 1;
			b++;
		}
		if((b == 4 || b == 6 || b == 9 || b == 11) && c == 31){
			c = 1;
			b++;
		}
		if(b == 2 && ((a % 100 == 0 && a % 400 == 0) || (a % 100 != 0 && a % 4 == 0)) && c == 30){
			c = 1;
			b++;
		}
		if(b == 2 && (!((a % 100 == 0 && a % 400 == 0) || (a % 100 != 0 && a % 4 == 0))) && c == 29){
			c = 1;
			b++;
		}
		if(b == 13){
			b = 1;
			a++;
		}
	}
	cout << sum << endl;
	return 0;
}

菜就多练

#include<bits/stdc++.h>
using namespace std;
int sy,sm,sd,ey,em,ed,ans,m[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int main()
{
	cin>>sy>>sm>>sd>>ey>>em>>ed;
	while(sy!=ey||sm!=em||sd!=ed){
	ans++;
	sd++;
	if(sy%4==0&&sy%100!=0||sy%400==0)m[2] = 29;
	else m[2] = 28;
	if(sd>m[sm]){
	sd = 1;
	sm++;
		}
	if(sm>12){
	sm = 1;
	sy++;
		}
	}
	cout<<ans;
	return 0;
}

#include #include #include #include #include #include #include #include #include<math.h> #include #include #include #include #include #include<stdio.h> #include #include #include #include<string.h> #include #include<bits/stdc++.h> using namespace std; #define in int #define ch char #define lo long #define fl float #define sh short #define db double #define str string #define ll long long #define ld long double #define lli long long int const int N =1e5+10; const int INF =0x3f3f3f3f; int lapeyear(int year2)//闰年判断函数 { if ((year2 % 400 == 0) || (year2 % 4 == 0 && year2 % 100 != 0)) return 1; else return 0; } int years1(int year1, int year2)//计算两年间的天数 { int d = 0, i, r = 0, z = 0; if (year1 < year2) { for (i = year1 + 1; i < year2; i++) { if (lapeyear(i))r++;//两年之间有多少个闰月 } d = (year2 - year1 - 1) * 365 + r; } else { for (i = year2 + 1; i < year1; i++) { if (lapeyear(i))r++;//两年之间有多少个闰月 } d = (year1 - year2 - 1) * 365 + r; } return d; } int months2(int year, int month, int date)//计算到年初日差的函数 { int i, d = 0; int years[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 }; if (lapeyear(year))years[1] = 29; for (i = 0; i < month - 1; i++) { d += years[i]; } d += date; return d; } int months1(int year, int month, int date)//计算到年底日差的函数 { int i, d = 0; int years[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 }; if (lapeyear(year))years[1] = 29; for (i = 11; i > month - 1; i--) { d += years[i]; } d = d + years[month - 1] - date; return d; } int main() { int year1, year2, month1, month2, date1, date2; int d = 0, i, m = 0; int x1, x2;//记录两个日期分别到年初和年底的天数 int years[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 }; cin >> year1 >> month1 >> date1 >> year2 >> month2 >> date2; if (year1 == year2)//大前提，同年时 { if (lapeyear(year1))years[1] = 29; if (month1 == month2)d = date1 > date2 ? date1 - date2 : d = date2 - date1;//同年同月 if (month1 != month2)//同年不同月 { //计算方法为求出两月之前的天数，然后求前一个月剩余的天数和后一个月已经开始的天数 if (month1 < month2) { { for (i = month1; i < month2 - 1; i++) { m += years[i]; } } d = years[month1 - 1] - date1; d += m + date2; } else { for (i = month2; i > month1 - 1; i++) { m += years[i]; } d = years[month2 - 1] - date2; d += m + date1; } } } else//求不同年的任意日期的天数差 { if (year1 > year2) { x1 = months1(year2, month2, date2);//x1为到年底的日差，x2为到年初的日差 x2 = months2(year1, month1, date1); d = years1(year2, year1); d += x1 + x2; } else { x2 = months2(year2, month2, date2); x1 = months1(year1, month1, date1); d = years1(year1, year2); d += x1 + x2; }

}
cout << d;
return 0;

}

/*********************************
备注：
*********************************/
#include<map>
#include<list>
#include<stack>
#include<queue>
#include<cmath>
#include<queue>
#include<stack>
#include<deque>
#include<math.h>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<fstream>
#include<stdio.h>
#include<sstream>
#include<iomanip>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<bits/stdc++.h>
using namespace std;
#define in int
#define ch char
#define lo long
#define fl float
#define sh short
#define db double
#define str string
#define ll long long
#define ld long double
#define lli long long int
const int N =1e5+10;
const int INF =0x3f3f3f3f;
int lapeyear(int year2)//闰年判断函数
{
	if ((year2 % 400 == 0) || (year2 % 4 == 0 && year2 % 100 != 0))
		return 1;
	else return 0;
}
int years1(int year1, int year2)//计算两年间的天数
{
	int d = 0, i, r = 0, z = 0;
	if (year1 < year2)
	{
		for (i = year1 + 1; i < year2; i++)
		{
			if (lapeyear(i))r++;//两年之间有多少个闰月
		}
		d = (year2 - year1 - 1) * 365 + r;
	}
	else
	{
		for (i = year2 + 1; i < year1; i++)
		{
			if (lapeyear(i))r++;//两年之间有多少个闰月
		}
		d = (year1 - year2 - 1) * 365 + r;
	}
	return d;
}
int months2(int year, int month, int date)//计算到年初日差的函数
{
	int i, d = 0;
	int years[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
	if (lapeyear(year))years[1] = 29;
	for (i = 0; i < month - 1; i++)
	{
		d += years[i];
	}
	d += date;
	return d;
}
int months1(int year, int month, int date)//计算到年底日差的函数
{
	int i, d = 0;
	int years[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
	if (lapeyear(year))years[1] = 29;
	for (i = 11; i > month - 1; i--)
	{
		d += years[i];
	}
	d = d + years[month - 1] - date;
	return d;
}
int main()
{
	int year1, year2, month1, month2, date1, date2;
	int d = 0, i, m = 0;
	int x1, x2;//记录两个日期分别到年初和年底的天数
	int years[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
	cin >> year1 >> month1 >> date1 >> year2 >> month2 >> date2;
	if (year1 == year2)//大前提，同年时
	{
		if (lapeyear(year1))years[1] = 29;
		if (month1 == month2)d = date1 > date2 ? date1 - date2 : d = date2 - date1;//同年同月
		if (month1 != month2)//同年不同月
		{
			//计算方法为求出两月之前的天数，然后求前一个月剩余的天数和后一个月已经开始的天数
			if (month1 < month2)
			{
				{
					for (i = month1; i < month2 - 1; i++)
					{
						m += years[i];
					}
				}
				d = years[month1 - 1] - date1;
				d += m + date2;
			}
			else
			{
					for (i = month2; i > month1 - 1; i++)
					{
						m += years[i];
		            }
				d = years[month2 - 1] - date2;
				d += m + date1;
			}
		}
	}
	else//求不同年的任意日期的天数差
	{
		if (year1 > year2)
		{
			x1 = months1(year2, month2, date2);//x1为到年底的日差，x2为到年初的日差
			x2 = months2(year1, month1, date1);
			d = years1(year2, year1);
			d += x1 + x2;
		}
		else
		{
			x2 = months2(year2, month2, date2);
			x1 = months1(year1, month1, date1);
			d = years1(year1, year2);
			d += x1 + x2;
		}

	}
	cout << d;
	return 0;
}