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日期差值

[编程题]日期差值
  • 热度指数:38669 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 64M,其他语言128M
  • 算法知识视频讲解
有两个日期,求两个日期之间的天数,如果两个日期是连续的我们规定他们之间的天数为两天

输入描述:
有多组数据,每组数据有两行,分别表示两个日期,形式为YYYYMMDD


输出描述:
每组数据输出一行,即日期差值
示例1

输入

20110412
20110422

输出

11
#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;
    }
}

编辑于 2020-04-23 17:35:03 回复(0)
直接计算两个日期之间的差值。
#include <iostream>
#include <cstdio>
using namespace std;
int months_day[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}};
int isLoop(int year){
    if((year%4==0 && year%100!=0) || year%400==0) return 1;
    else return 0;
}
int main(){
    int year1,month1,day1;
    int year,month,day;
    while(scanf("%4d%2d%2d",&year,&month,&day)!=EOF){
        int sum_days=0;
        scanf("%4d%2d%2d",&year1,&month1,&day1);
        if(year==year1 && month==month1) sum_days=abs(day-day1)+1;
        else if(year==year1){
            if(month<month1){
                for(int i=month;i<=month1;i++){
                    if(i==month) sum_days+=(months_day[month][isLoop(year)]-day+1);
                    else if(i==month1) sum_days+=(day1);
                    else sum_days+=months_day[i][isLoop(year)];
                }
            }else{
                for(int i=month1;i<=month;i++){
                    if(i==month1) sum_days+=(months_day[month1][isLoop(year)]-day1+1);
                    else if(i==month) sum_days+=(day);
                    else sum_days+=months_day[i][isLoop(year)];
                }
            }
        }//elseif
        else{
           if(year<year1){
               for(int i=year;i<=year1;i++){
                   if(i==year){
                       for(int j=month;j<=12;j++){
                           if(j==month) sum_days+=(months_day[j][isLoop(i)]-day+1);
                           else sum_days+=(months_day[j][isLoop(i)]);
                       } 
                   }
                   else if(i==year1){
                       for(int j=1;j<=month1;j++){
                           if(j==month1) sum_days+=(day1);
                           else sum_days+=(months_day[j][isLoop(i)]);
                       }
                   }
                   else{
                       if(isLoop(i)==1) sum_days+=366;
                       else sum_days+=365;
                   }
               }
           }
           else{
               for(int i=year1;i<=year;i++){
                   if(i==year1){
                       for(int j=month1;j<=12;j++){
                           if(j==month1) sum_days+=(months_day[j][isLoop(i)]-day1+1);
                           else sum_days+=(months_day[j][isLoop(i)]);
                       } 
                   }
                   else if(i==year){
                       for(int j=1;j<=month;j++){
                           if(j==month) sum_days+=(day);
                           else sum_days+=(months_day[j][isLoop(i)]);
                       }
                   }
                   else{
                       if(isLoop(i)==1) sum_days+=366;
                       else sum_days+=365;
                   }
               }
           }
        }//else
        cout<<sum_days<<endl;
    }//while
    return 0;
}


发表于 2022-02-05 00:07:02 回复(1)
记录下垃圾题解
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;
    }
}


发表于 2021-06-10 11:55:02 回复(0)
Java
import java.time.LocalDate;
import java.time.temporal.ChronoUnit;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNext()){
            LocalDate date1 = getDate(scanner.next());
            LocalDate date2 = getDate(scanner.next());
            System.out.println(date1.until(date2, ChronoUnit.DAYS)+1);
        }
    }

    static LocalDate getDate(String s){
        String year = s.substring(0, 4);
        String month = s.substring(4, 6);
        String day = s.substring(6);
        return LocalDate.of(Integer.parseInt(year),Integer.parseInt(month),Integer.parseInt(day));
    }
}


发表于 2020-03-18 17:40:20 回复(0)
 #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;
    
}

发表于 2020-02-15 12:48:46 回复(0)
写的有点复杂,其实思路很简单,但是中间的各种判断相加啥的写的有点复杂。。。献丑了
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
int i,j,sum;
int m0[12]={31,28,31,30,31,30,31,31,30,31,30,31};
int m1[12]={31,29,31,30,31,30,31,31,30,31,30,31};
int a[8];

int test(int year)//判断闰年
{
    if(year%400==0)
        return 1;
    else
    {
        if(year%4==0 && year%100!=0)
            return 1;
        else
            return 0;
    }
}

void get_info(int &date,int &year,int &mon,int &day)//从输入分别获取年月日
{
    memset(a,0,sizeof(a));
    i=10000000;
    j=0;
    while(date>0)
    {
        a[j]=date/i;
        if(a[j]>=0)
        {
            date-=a[j]*i;
            j++;
        }
        i/=10;
    }
    year=a[0]*1000+a[1]*100+a[2]*10+a[3];
    mon=a[4]*10+a[5];
    day=a[6]*10+a[7];
}

void count_day(int year1,int mon1,int day1,int year2,int mon2,int day2)//当输入不是同一年时获取结果
{
    sum=1;
    for(i=year1+1;i<year2;i++)
    {
        if(test(i)==1)
            sum+=366;
        else
            sum+=365;
    }
    if(test(year1)==1)
    {
        for(i=mon1+1;i<=12;i++)
            sum+=m1[i-1];
        sum=sum+m1[mon1-1]-day1;
    }
    else{
        for(i=mon1+1;i<=12;i++)
            sum+=m0[i-1];
        sum=sum+m0[mon1-1]-day1;
    }

    if(test(year2)==1)
    {
        for(i=1;i<mon2;i++)
            sum+=m1[i-1];
        sum=sum+day2;
    }
    else{
        for(i=1;i<mon2;i++)
            sum+=m0[i-1];
        sum=sum+day2;
    }
    cout<<sum<<endl;
}
void count_day_same(int year,int mon1,int day1,int mon2,int day2)//当输入是同一年时获取结果
{
    if(test(year)==1)
    {
        if(mon1<mon2)
        {
            for(i=mon1+1;i<mon2;i++)
                sum+=m1[i-1];
            sum+=m1[mon1-1]-day1+1;
            sum+=day2;
        }
        else if(mon1>mon2)
        {
            for(i=mon2+1;i<mon1;i++)
                sum+=m1[i-1];
            sum+=m1[mon2-1]-day2+1;
            sum+=day1;
        }
        else
            sum=day1>day2?(day1-day2+1):(day2-day1+1);
    }
    else{
        if(mon1<mon2)
        {
            for(i=mon1+1;i<mon2;i++)
                sum+=m0[i-1];
            sum+=m0[mon1-1]-day1+1;
            sum+=day2;
        }
        else if(mon1>mon2)
        {
            for(i=mon2+1;i<mon1;i++)
                sum+=m0[i-1];
            sum+=m0[mon2-1]-day2+1;
            sum+=day1;
        }
        else
            sum=day1>day2?(day1-day2+1):(day2-day1+1);
    }
    cout<<sum<<endl;
}

int main(void)
{
    int date1,date2;
    int year1,year2,mon1,mon2,day1,day2;
    while(cin>>date1>>date2)
    {
        get_info(date1,year1,mon1,day1);
        get_info(date2,year2,mon2,day2);
        int s=day1>day2?(day1-day2):(day2-day1);
        if(year1==year2&&mon1==mon2&&s==1)
        {
            cout<<"2"<<endl;
            break;
        }
        else if(year1<year2)
            count_day(year1,mon1,day1,year2,mon2,day2);
        else if(year1>year2)
            count_day(year2,mon2,day2,year1,mon1,day1);
        else
            count_day_same(year1,mon1,day1,mon2,day2);
    }
    return 0;
}
发表于 2019-11-24 15:04:07 回复(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;
}
编辑于 2019-05-14 17:14:45 回复(0)
细节题!!
#include <stdio.h>
#include <math.h>
char date1[10],date2[10];
int month[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
struct DATE{
    int y;
    int m;
    int d;
} dt1,dt2,tmp;
int is_LeapYear(int year){
    return ((year%4==0 && year%100!=0) || (year%400==0))?1:0;
}
int main(){
    int i,res;
    while(scanf("%s",date1)!=EOF){
        scanf("%s",date2);
        dt1.y=dt1.m=dt1.d=0;
        dt2.y=dt2.m=dt2.d=0;
        for(i=0;i<4;i++){
            dt1.y+=((date1[i]-'0')*pow(10,3-i));
            dt2.y+=((date2[i]-'0')*pow(10,3-i));
        }
        for(i=4;i<6;i++){
            dt1.m+=((date1[i]-'0')*pow(10,5-i));
            dt2.m+=((date2[i]-'0')*pow(10,5-i));
        }
        for(i=6;i<8;i++){
            dt1.d+=((date1[i]-'0')*pow(10,7-i));
            dt2.d+=((date2[i]-'0')*pow(10,7-i));
        }
        if((dt1.y>dt2.y) || (dt1.y==dt2.y && dt1.m>dt2.m)
            || (dt1.y==dt2.y && dt1.m==dt2.m && dt1.d>dt2.d)){
            tmp=dt1;
            dt1=dt2;
            dt2=tmp;
        }
        //年月相同
        if(dt1.y==dt2.y && dt1.m==dt2.m)
            res=dt2.d-dt1.d+1;
        //年相同
        else if(dt1.y==dt2.y){
            res=0;
            if(is_LeapYear(dt1.y))
                month[2]=29;
            else
                month[2]=28;
            for(i=dt1.m+1;i<dt2.m;i++)
                res+=month[i];
            res+=(month[dt1.m]-dt1.d+1);
            res+=(dt2.d);
        }
        //年不同
        else{
            res=0;
            for(i=dt1.y+1;i<dt2.y;i++){
                if(is_LeapYear(i))
                    res+=366;
                else
                    res+=365;
            }
            if(is_LeapYear(dt1.y))
                month[2]=29;
            else
                month[2]=28;
            for(i=dt1.m+1;i<=12;i++)
                res+=month[i];
            res+=(month[dt1.m]-dt1.d+1);
            if(is_LeapYear(dt2.y))
                month[2]=29;
            else
                month[2]=28;
            for(i=1;i<dt2.m;i++)
                res+=month[i];
            res+=(dt2.d);
        }
        printf("%d\n",res);
    }
    return 0;
}

发表于 2019-01-04 09:39:04 回复(0)
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();
    }
    
}

发表于 2018-09-19 16:04:49 回复(0)
//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;
}

发表于 2018-02-26 04:03:58 回复(0)
#include<stdio.h>
#define isyeap(x) x%100!=0&&x%4==0||x%400==0?1:0
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++;
        }
    }
    }
};
int buf[5001][13][32];
int abs(int x)
{
    return x<0?-x:x;
}
int main()
{
    Date tmp;
    int cnt=0;
    tmp.day=1;
    tmp.month=1;
    tmp.year=0;
    while(tmp.year!=5001)
    {
        buf[tmp.year][tmp.month][tmp.day]=cnt;
        tmp.nextDay();
        cnt++;
    }
    int d1,m1,y1;
    int d2,m2,y2;
    while(scanf("%4d%2d%2d",&y1,&m1,&d1)!=EOF)
    {
        scanf("%4d%2d%2d",&y2,&m2,&d2);
        printf("%d\n",abs(buf[y2][m2][d2]-buf[y1][m1][d1])+1);
    }
}
 
发表于 2018-01-03 18:54:30 回复(0)
#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;
}

发表于 2018-03-10 15:36:09 回复(3)

//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;
}
编辑于 2019-01-23 17:35:37 回复(10)

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
编辑于 2018-01-27 16:12:49 回复(3)
我用cin输入数值编译不过去,用scanf输入数值就可以编,操蛋,这是为什么?
发表于 2022-02-12 19:41:42 回复(0)
tips:1.计算两日期距元年元旦的天数相减再+1
       2.计算天数用bool值和2维数组感觉很直观
       3.输入格式用c的scanf很清晰
#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;
}

发表于 2020-02-17 01:14:24 回复(2)
#include <iostream>

using namespace std;

int md[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

// 日期结构体
struct Date {
    int y;
    int m;
    int d;
};

// 判断是否为闰年
bool isRun(int y) {
    return (y%4==0&&y%100!=0||y%400==0);
}

// 获取指定年份的月份的天数
int get_md(int y, int m) {
    if(isRun(y) && m == 2) {
        return 29;
    } else {
        return md[m];
    }
}

// 转化日期格式
Date get_date(int dateNum) {
    Date date;
    date.y = dateNum/10000;
    date.m = dateNum/100%100;
    date.d = dateNum%100;
    return date;
}

// 下一天
Date next_day(Date date) {
    date.d += 1;
    if(date.d > get_md(date.y, date.m)) {
        date.m += 1;
        date.d = 1;
    }
    if(date.m > 12) {
        date.y += 1;
        date.m = 1;
        date.d = 1;
    }
    return date;
}

// 判断两个日期是否相等
bool equal(Date date1, Date date2) {
    return date1.y == date2.y && date1.m == date2.m && date1.d == date2.d;
}

int main()
{
    int dateNum1, dateNum2;
    while(cin>>dateNum1>>dateNum2) {
        int count = 1;
        Date date1 = get_date(dateNum1);
        Date date2 = get_date(dateNum2);
        while(!equal(date1, date2)) {
            count++;
            date1 = next_day(date1);
        }
        cout<<count<<endl;
    }
    return 0;
}

发表于 2018-03-24 15:04:53 回复(0)

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
发表于 2019-01-07 11:24:37 回复(0)
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

int daytab[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 isleapyear(int x){
    if(x%4==0&&x%100!=0||x%400==0)
    return true;
    else return false;
}

int numberofyear(int x){
    if(isleapyear(x))
    return 366;
    else return 365;
}

char a[9],b[9];
int main(){
    while(cin>>a>>b){
        int year,month,day;
        int year1,month1,day1;
        int num1=0,num2=0;
        sscanf(a,"%4d%2d%2d",&year,&month,&day);
        sscanf(b,"%4d%2d%2d",&year1,month1,day1);
        for(int i=0;i<year;i++){
            num1+=numberofyear(i);
        }
        int row=isleapyear(year);
        for(int i=0;i<month;i++){
            num1+=daytab[row][i];
        }
        num1+=day;
        for(int i=0;i<year1;i++){
            num2+=numberofyear(i);
        }
        row=isleapyear(year1);
        for(int i=0;i<month1;i++){
            num2+=daytab[row][i];
        }
        num2+=day1;
        int num=abs(num2-num1)+1;
        cout<<num<<endl;    
    }
    return 0;    
}
发表于 2024-01-26 15:47:30 回复(0)
将大佬的代码自己写了一遍
#include<iostream>
#include<stdio.h>
#include<string>
#include<cmath>

using namespace std;

//这里使用计算两个日期与0000 00 00 之间的距离distance1和distance2,最终返回abs(distance1-distance2)+1即可

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

int monthday[2][12] = {
    {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
    {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
};


int main() {
    char daytime1[9];
    char daytime2[9]; //表示输入的两个日期

    while (cin >> daytime1 >> daytime2) {
        int year1, month1, day1;
        int year2, month2, day2;
sscanf(daytime1,"%4d%2d%2d",&year1,&month1,&day1);
sscanf(daytime2,"%4d%2d%2d",&year2,&month2,&day2);

        int distance1 = 0, distance2 = 0;

        for (int i = 1; i < year1; i++) {
            distance1 += isleapyear(i) ? 366: 365;
        }

        for (int i = 1; i < month1; i++) {
            distance1 += monthday[isleapyear(year1)][i-1];
        }

        distance1 += day1;

        for (int i = 1; i < year2; i++) {
            distance2 += isleapyear(i) ? 366: 365;
        }

        for (int i = 1; i < month2; i++) {
            distance2 += monthday[isleapyear(year1)][i-1];
        }

        distance2 += day2;

        int res = 0;
        res = abs(distance1 - distance2) + 1;
        cout << res << endl;
    }
}


发表于 2024-01-18 13:54:08 回复(0)