#include<stdio.h>
#define leap(x) (x%400==0||(x%4==0&&x%100!=0)?1:0)
#define Abs(x) ((x)>0?(x):-(x))
typedef struct date
{
    int y, m, d;
}Date;
const int yd[2][13]={{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
                     {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};
                     
int days(Date x)
{
    int sum=0, i;
    int y=x.y, m=x.m, d=x.d;
    for(i=0; i<y; i++)
    {
        if(leap(i))
        {
            sum+=366;
        }
        else
        {
            sum+=365;
        }
    }
    for(i=1; i<m; i++)
    {
        sum+=yd[leap(y)][i];
    }
    sum+=d;
    return sum;
}
int main(void)
{
    Date a, b;
    while(scanf("%4d%2d%2d", &a.y, &a.m, &a.d)!=EOF)
    {
        scanf("%4d%2d%2d", &b.y, &b.m, &b.d);
        printf("%d\n", Abs(days(a)-days(b))+1);
    }
    return 0;
}

//计算两个日期距离0000年0月1日的天数再求差

#include <bits/stdc++.h>
using namespace std;
int mon[12]={0,31,59,90,120,151,181,212,243,273,304,334};//字典 
int cal(int y,int m,int d)//给出年月日，计算距离0000年0月1日的天数和
{
    return y*365+y/4-y/100+y/400+mon[m-1]+d-1+(y/100!=0&&y/4==0||y/400==0&&m>2);
}
int main()
{
    int x[2],year,month,day;
    for(int k=0;k<2;k++)//循环两次读入两个日期
    {
        scanf("%4d%2d%2d",&year,&month,&day);
        x[k]=cal(year,month,day);
    }
    cout<<abs(x[0]-x[1])+1;
}

python只需要一行代码搞定。

要借助datetime库。

from datetime import datetime
while True:
    try:
        print(-(datetime.strptime(input(),"%Y%m%d")-datetime.strptime(input(),"%Y%m%d")).days+1)
    except:
        break

#include<bits/stdc++.h>
using namespace std;
const int month[2][13] = { {0,31,28,31,30,31,30,31,31,30,31,30,31},
			        {0,31,29,31,30,31,30,31,31,30,31,30,31 } };
bool leapyear(int y) {
	if (y % 4 == 0 && y % 100 != 0 || y % 400 == 0)
		return true;
	else return false;
}
int numofyear(int y) {
	if (leapyear(y))
		return 366;
	else return 365;
}
int sumofdate(int y, int m, int d) {
	int sum = 0;
	for (int i = 0; i < y; i++) {
		sum += numofyear(i);
	}
	for (int i = 0; i < m; i++) {
		sum += month[leapyear(y)][i];
	}
	sum += d;
	return sum;
}
int main() {
	int y1, m1, d1, y2, m2, d2;
	while (scanf("%04d%02d%02d", &y1, &m1, &d1) != EOF) {
		scanf("%04d%02d%02d", &y2, &m2, &d2);
		printf("%d\n", abs(sumofdate(y1, m1, d1)-sumofdate(y2, m2, d2))+1);
	}
	return 0;
}

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;
int tab[2][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31},
                {0,31,29,31,30,31,30,31,31,30,31,30,31}};
bool leapyear(int year)
{
    return (year%4==0&&year%100!=0)||(year%400==0);
}
int numberofyear(int year)
{
    if(leapyear(year))
        return 366;
    else return 365;
}
int main()
{
    char s1[9];
    char s2[9];
    while(cin>>s1>>s2)
    {
        int year1,year2,month1,month2,day1,day2;
        int num1=0;
        int num2=0;
        sscanf(s1,"%4d%2d%2d",&year1,&month1,&day1);
        sscanf(s2,"%4d%2d%2d",&year2,&month2,&day2);
        for(int i=0;i<year1;i++)
        {
            num1+=numberofyear(i);
        }
        for(int j=0;j<month1;j++)
        {
            num1+=tab[leapyear(year1)][j];
        }
        num1+=day1;
        for(int i=0;i<year2;i++)
        {
            num2+=numberofyear(i);
        }
        for(int j=0;j<month2;j++)
        {
            num2+=tab[leapyear(year2)][j];
        }
        num2+=day2;
        int num=abs(num2-num1)+1;
        cout<<num<<endl;
    }
}

using namespace std;
#include <iostream>
#include <cstring>
#include <sstream>
int sum[13] = {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365};
bool judge(int y){
    return (y%4 == 0&nbs***bsp;y%400 == 0) and y%100 != 0;
}
template<class out_type, class in_value>
out_type convert(const in_value & t)
{
    stringstream stream;
    stream << t;
    out_type result;
    stream >> result;
    return result;
}

int Div(int y, int x){
    int ans(0);
    if(abs(y-x) >= 1){
        int low = min(y, x), high = max(y, x);
        while(low < high){
            if(judge(low))ans += 366;
            else ans += 365;
            ++low;
        }
    }
    return ans;
}
int count(int y,int m, int d){
    int ans(0);
    if(judge(y) and m>2)ans++;
    ans += sum[m-1] + d;
    return ans;
}
void change(int &y, int &m, int &d,string n){
    y = convert<int>(n.substr(0, 4));
    m = convert<int>(n.substr(4, 2));
    d = convert<int>(n.substr(6, 2));
}
int main(){
    string n1, n2;
    while(cin>>n1>>n2){
        int y1, y2, m1, m2, d1, d2, c1, c2;
        change(y1, m1, d1, n1);
        change(y2, m2, d2, n2);
        c1 = count(y1, m1, d1);
        c2 = count(y2, m2, d2);
        cout<<Div(y1, y2) + c2 - c1 + 1;
    }
}

 #include<stdio.h>
#define isyear(x) x%10!=0&&x%4==0||x%400==0?1:0
int dom[13][2]
{
0,0,
31,31,
28,29,
31,31,
30,30,
31,31,
30,30,
31,31,
31,31,
30,30,
31,31,
30,30,
31,31,
};
typedef struct data
{
    int year;
    int month;
    int day;
    void addday()
    {
        day++;
        if(day>dom[month][isyear(year)])
        {
            day=1;
            month++;
            if(month>12)
            {
                month=1;
                year++;
            }
        }
    }
}data;
int buf[5000][13][31]={0};
int ABS(int a){return a<0?-a:a;}
int main()
{
     int cnt=0;
    data tem;
    tem.year=0;
    tem.month=1;
    tem.day=1;
    while(tem.year!=5000)
    {
        
        tem.addday();
        cnt++;
        buf[tem.year][tem.month][tem.day]=cnt;
    }
    int y1,m1,d1;
    int y2,m2,d2;
    while(scanf("%4d%2d%2d",&y1,&m1,&d1)!=EOF)
          {
             scanf("%4d%2d%2d",&y2,&m2,&d2);
             printf("%d",ABS(buf[y1][m1][d1]-buf[y2][m2][d2])+1);    
          }
    return 0;
    
}

思路:

将小的日期记为A，大的日期记为B

计算 A年的 1.1 (A11) 到B年1.1 (B11)的天数 A11toB11 (记得计算闰年)

计算 A到本年的日期 Ato11

计算 B到本年的日期 Bto11

日期差值为

A11toB11 + Bto11 - Ato11

代码如下：

#include
#include
#include 
using namespace std; 
int beginDayOfMonth[13][2] = {
    0, 0,
    0, 0,
    31, 31,
    59, 60,
    90, 91,
    120, 121,
    151, 152,
    181, 182,
    212, 213,
    243, 244,
    273, 274,
    304, 305,
    334, 335 
};
int isRunYear(int year)
{
    return year%100 != 0 && year%4 == 0 || year%400 == 0 ? 1:0;
} 
int dayto11(int year, int month, int day)
{
    int isRun = isRunYear(year);
    if(isRun)
    {
         return day + beginDayOfMonth[month][1];
    }
    else return day + beginDayOfMonth[month][0];
    // return day + beginDayOfMonth[month][isRun];  但这里需要保证 isRun 只取 01 
} 
int main()
{
    int AtoA11, BtoB11, A11toB11;
    int y1, m1, d1;
    int y2, m2, d2;
    int disDay;
    while(scanf("%4d%2d%2d", &y1, &m1, &d1) != EOF)
    {
        scanf("%4d%2d%2d", &y2, &m2, &d2);
        if(y1 > y2 || (y1 == y2 && m1 > m2) || (y1 == y2 && m1 == m2 && d1 > d2))
        {
            int tmp = y1;
            y1 = y2;
            y2 = tmp;
            tmp = m1;
            m1 = m2;
            m2 = tmp;
            tmp = d1;
            d1 = d2;
            d2 = tmp;    
        }
        AtoA11 = dayto11(y1, m1, d1);
        BtoB11 = dayto11(y2, m2, d2);
        // cout << AtoA11 << endl;
        // cout << BtoB11 << endl;
        A11toB11 = 0;
        for(int i = y1; i < y2; i++)
        {
            if(isRunYear(i)) A11toB11 += 366;
            else A11toB11 += 365;
        }
        disDay = A11toB11 + BtoB11 - AtoA11 + 1;  // 这里是 把 最后一天的也给算上 
        printf("%d\n", disDay);     
    }
    return 0;
}

思路二
主要参考“看不见的风景201803281904”。
这里我对计算距离0年1月1日的方法进行了修改，

#include<cstdio>
#include<iostream> 
#include<cmath> 
using namespace std;
int mon[12] = {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334};

int sumRun1(int y)
{
    // 计算[0, y]之间 闰年 的次数 
    // 若 y在 0000 年之前, 则为负数，记录了[y, -1]闰年的次数(的相反数) 
    return 1 + floor(double(y-1)/4+1e-4) - floor(double(y-1)/100+1e-4)  + floor(double(y-1)/400+1e-4);
    // y-1是因为如果 第y年是闰年，那么会对 00000101至 y+1年1月1日 的天数造成影响 
    // +1 是为了 将0000年记为闰年 
    // floor(double(y-1)/4+1e-4) 是为了 计算在公元前的闰年 
    // + 1e-4 是为了避免浮点误差

    /* 
    其实这段代码的效果就是为了对应
    年数 
        -5 -4 -3 -2 -1  0  1  2  3  4  5  6  
    累计闰年个数
        -1 -1  0  0  0  0  1  1  1  1  2  2 

    */ 
} 
int isRunYear(int year)
{
    return year%100 != 0 && year%4 == 0 || year%400 == 0 ? 1:0;
} 
int sumRun2(int y) // 和这个效果是一样的
{
    int numOfRun = 0;
    if(y < 0)
    {
        for(int i = y; i < 0; i++)
        {
            if(isRunYear(i)) numOfRun--;
        }    
    }
    else
    {
        for(int i = 0; i < y; i++)
        {
            if(isRunYear(i)) numOfRun++;
        }    
    } 

    return numOfRun;
}
// 给出年月日，计算距离0000年 1 月 1 日的 天数 
int calto01(int y, int m, int d)
{
    return sumRun1(y) + y * 365 + mon[m-1] + d-1 + (((y%100 != 0 && y%4 == 0) || y%400 == 0) && m > 2); 
}

int main()
{
    int disto01[2], year, month, day;
    int y0, m0, d0;
    int y1, m1, d1;
    while(scanf("%4d%2d%2d", &y0, &m0, &d0) != EOF)
    {
        scanf("%4d%2d%2d", &y1, &m1, &d1);
        disto01[0] = calto01(y0, m0, d0);
        disto01[1] = calto01(y1, m1, d1);
        // cout << disto01[0] << " " << disto01[1] << endl; 
        printf("%d\n", abs(disto01[0] - disto01[1]) + 1);
    }
    return 0;
}

#include<iostream>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;

int bijiao(int y1,int m1,int d1,int y2,int m2,int d2){  //比较日期大小
    if(y1>y2)
        return 1;
    else if((y1==y2)&&(m1>m2))
        return 1;
    else if((y1==y2)&&(m1==m2)&&(d1>d2))
        return 1;
    return 0;
}

int runnian(int a){ //判断是否是闰年
    if((a%400==0)||(a%100!=0&&a%4==0))
        return 1;
    return 0;
}

int main(){
    long int a,b;
    cin>>a>>b;
    int s1[12]={31,28,31,30,31,30,31,31,30,31,30,31};
    int s2[12]={31,29,31,30,31,30,31,31,30,31,30,31};
    int y1=a/10000;
    int y2=b/10000;
    int m1=a/100%100;
    int m2=b/100%100;
    int d1=a%100;
    int d2=b%100;
    int k=0;
    if(bijiao(y1,m1,d1,y2,m2,d2)==0){   //总是让y1对应的时间是最晚的
        swap(y1,y2);
        swap(m1,m2);
        swap(d1,d2);
    }
    while(y1!=y2||m1!=m2||d1!=d2){  //一天一天往上增加，直到等于最晚的时间
        if(runnian(y2)){    //若是闰年则使用数组s2，否则s1
            if((d2+1)>s2[m2-1]){    //d2增加一天，判断是否超出月份
                if(m2+1>12){    //超出月份则月份加一，判断是否过了12月
                    y2++;   //超过12月则年份加一，时间归1
                    m2=1;
                    d2=1;
                }
                else {  //若没有超过年份，则月份加1，天数变为1；
                    m2++;
                    d2=1;
                }
            }
        else {  //若是没有超过月份，则天数加一
            d2++;
        }
    }
    else {  //不是闰年，进行相同操作
        if((d2+1)>s1[m2-1]){
                if(m2+1>12){
                    y2++;
                    m2=1;
                    d2=1;
                }
                else {
                    m2++;
                    d2=1;
                }
            }
        else {
            d2++;
        }
    }
    k++;//每次天数加一后，K++
    }
    cout<<++k; //最后输出要增加一天
}

	

分别对年月日进行差值计算累加天数

#coding : utf-8

year = [365, 366]
month = [[31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31],[31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]]
def isRun(y):
    if y%400==0 or (y%100!=0 and y%4==0):
        return 1
    return 0
def count(y1, m1, d1, y2, m2, d2):
    cnt = 0
    for y in xrange(y1, y2):
        cnt+=year[isRun(y)]
    y2_isRun = isRun(y2)
    for m in xrange(m1, m2):
        cnt+=month[y2_isRun][m-1]
    return cnt+d2-d1+1
while True:
    try:
        date1 = raw_input().strip()
        date2 = raw_input().strip()
        y1 = int(date1[0:4])
        y2 = int(date2[0:4])
        m1 = int(date1[4:6])
        m2 = int(date2[4:6])
        d1 = int(date1[6:8])
        d2 = int(date2[6:8])
        if(y1>y2) or (y1==y2 and m1>m2) or (y1==y2 and m1==m2 and d1>d2):
            y1 = int(date2[0:4])
            y2 = int(date1[0:4])
            m1 = int(date2[4:6])
            m2 = int(date1[4:6])
            d1 = int(date2[6:8])
            d2 = int(date1[6:8])
        print count(y1, m1, d1, y2, m2, d2)
    except:
        break

import datetime
while True:
    try:
        date1 = input()
        date2 = input()
        print((datetime.datetime(int(date2[:4]),int(date2[4:6]),int(date2[6:]))-datetime.datetime(int(date1[:4]),int(date1[4:6]),int(date1[6:]))).days+1)
    except Exception:
        break

import java.io.BufferedInputStream;
import java.time.LocalDate;
import java.time.format.DateTimeFormatter;
import java.time.temporal.ChronoUnit;
import java.util.Scanner;

public class Main
{
    public static void main(String... as)
    {
        Scanner sc = new Scanner(new BufferedInputStream(System.in));
        LocalDate ld1 = LocalDate.parse(sc.next(), DateTimeFormatter.ofPattern("yyyyMMdd"));
        LocalDate ld2 = LocalDate.parse(sc.next(), DateTimeFormatter.ofPattern("yyyyMMdd"));
        System.out.println(ld1.until(ld2, ChronoUnit.DAYS) + 1);
        sc.close();
    }
    
}

	

//因为还不会格式输入，所以输入年月日的时候用了笨办法，希望大佬能帮忙改一下，谢谢！
#include<iostream>

#define ISYEAP(x) ((x%4==0&&x%100!=0)||x%400==0) ?1:0



using namespace std;

int dayofmonth[13][2] = {

    0,0,

    31,31,28,29,31,31,30,30,31,31,30,30,

    31,31,31,31,30,30,31,31,30,30,31,31

};

struct  Date {

    int day;

    int month;

    int year;

    void nextday() {

        day++;

        if (day>dayofmonth[month][ISYEAP(year)])

        {

            day = 1;

            month++;

            if (month>12)

            {

                month = 1;

                year++;

            }//if

        }//if

    }//nextday

};//data



int buf[5001][12][32];



int abs(int x) {

    return x<0 ? -x : x;

}

int main() {

    Date tmp;

    tmp.year = 0;

    tmp.month = 1;

    tmp.day = 1;

    int num = 0;

    while (tmp.year < 5001) {

        buf[tmp.year][tmp.month][tmp.day] = num;

        tmp.nextday();

        num++;

    }//while

    int d1, m1, y1, temp1;

    int d2, m2, y2, temp2;

    while (cin >> temp1) {

        d1 = temp1 % 100; temp1 = temp1 / 100;

        m1 = temp1 % 100; y1 = temp1 / 100;

        cin >> temp2;

        d2 = temp2 % 100; temp2 = temp2 / 100;

        m2 = temp2 % 100; y2 = temp2 / 100;

        cout << abs(buf[y2][m2][d2] - buf[y1][m1][d1])+1<< endl;

    }

    return 0;

}

献丑了。

#include<cstdio>
struct Time {
    int y;
    int m;
    int d;
}t1,t2;
bool Is366(int year){
    return (year % 4 == 0 && year % 100 != 0 || year % 100 == 0 && year % 400 == 0);
}
int month(int year, int month){
    int Is31[7] = { 1,3,5,7,8,10,12 };
    for (int i = 0; i < 7; i++){
        if (Is31[i] == month)
            return 31;
    }
    if (month == 2){
        if (Is366(year))
            return 29;
        else return 28;
    }
    else return 30;
}
void swap(struct Time *a, struct Time *b) {
    struct Time temp;
    temp=*a;
    *a = *b;
    *b = temp;
}
int main(){    
    int val;
    while (scanf("%4d%2d%2d %4d%2d%2d", &t1.y, &t1.m, &t1.d, &t2.y, &t2.m, &t2.d) != EOF) {
        if (t1.y > t2.y) swap(&t1, &t2);
        else if(t1.y==t2.y){
            if (t1.m > t2.m) swap(&t1, &t2);
            else if(t1.m==t2.m)
                if(t1.d>t2.d) swap(&t1, &t2);
        }
        if (t1.y == t2.y) {
            if (t1.m == t2.m)
                val = t2.d - t1.d + 1;
            else {
                val = month(t1.y, t1.m) - t1.d + 1 + t2.d;
                for (int i = t1.m + 1; i <= t2.m - 1; i++)
                    val += month(t1.y, i);
            }
        }
        else {
            val = month(t1.y, t1.m) - t1.d + 1;
            for (int i = t1.m + 1; i <= 12; i++)
                val += month(t1.y, i);
            for (int i = t1.y + 1; i < t2.y; i++) {
                if (Is366(i)) val += 366;
                else val += 365;
            }
            for (int i = 1; i < t2.m; i++)
                val += month(t2.y, i);
            val += t2.d;
        }
        printf("%d\n", val);
    }
    return 0;
}