[编程题]日期差值
#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(2)
//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(6)
#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;
}
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)
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(1)
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];
{
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)
献丑了，用了最笨的方法了，一天一天数过去的。。。
思路：从较小的日期开始，一天一天往上增加，直到等于较大日期为止。
#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
}
编辑于 2019-03-12 20:53:19
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)
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;
}
#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)
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;
}
编辑于 2018-01-06 15:54:02
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);
}
}
#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)
原谅我不会输入数据格式控制
#include<iostream>
(720)#include<vector>
#include<cmath>
(808)#include<string>
#include<algorithm>
(831)#include<map>
#include<regex>
using namespace std;
bool isRun(int n)
{
if (n % 400 == 0 || (n % 4 == 0 && n % 100 != 0)) return true;
return false;
}
int main()
{
string x, y;
int a[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
while (cin >> x >> y)
{
int year1 = atoi(x.substr(0, 4).c_str());
int month1 = atoi(x.substr(4, 2).c_str());
int day1 = atoi(x.substr(6).c_str());
int year2 = atoi(y.substr(0,4).c_str());
int month2 = atoi(y.substr(4, 2).c_str());
int day2 = atoi(y.substr(6).c_str());
int ans1 = 365*(year1-1);
int ans2 = 365 * (year2 - 1);
int maxs = max(year1, year2);
for (int i = 1; i < maxs; i++) //0001-01-01
{
if (isRun(i))
{
if (i < year1) ans1++;
if (i <year2) ans2++;
}
}
if (isRun(year1)) a[2] = 29;
else a[2] = 28;
for (int i = 1; i < month1; i++)
ans1 += a[i];
ans1 += day1;
if (isRun(year2)) a[2] = 29;
else a[2] = 28;
for (int i = 1; i < month2; i++)
ans2 += a[i];
ans2 += day2;
cout << abs(ans1 - ans2)+1 << endl;
}
}
发表于 2020-04-01 00:08:55
回复(0)
#include<stdio.h>
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}
};
int IsLeapYear(int year){
return (year % 4 == 0&& year %100 != 0) || (year % 400 == 0);
}
int NumberOfYear(int year){
if(IsLeapYear(year)){
return 366;
}else{
return 365;
}
}
int main(){
int date_1,date_2;
int year_1,year_2,month_1,month_2,day_1,day_2,number = 0;
int big,small;
while(scanf("%d %d",&date_1,&date_2) != EOF){
big = date_1>=date_2?date_1:date_2;
small = date_1<=date_2?date_1:date_2;
year_1 = big / 10000;
year_2 = small / 10000;
month_1 = (big - year_1*10000) / 100;
month_2 = (small - year_2*10000) / 100;
day_1 = big-year_1*10000-month_1*100;
day_2 = small-year_2*10000-month_2*100;
if(year_1 == year_2){
//判断是否闰年
int row = IsLeapYear(year_1);
for(int i = month_2;i<month_1;i++)
number+=daytab[row][i];
number = number-day_2+day_1+1;
printf("%d\n",number);
}
else{
//分为三部分
//第一部分为小日子是第多少天，并用小日子的一年天数减去小日子是第多少天
int number_1 = 0,number_2 = 0;
int row_2 = IsLeapYear(year_2);
for(int i = 1;i<month_2;i++){
number_2 += daytab[row_2][i];
}
number = NumberOfYear(year_2)-number_2-day_2;
//第二部分为大日子是第多少天，直接进行相加
int row_1 = IsLeapYear(year_1);
for(int i = 1;i<month_1;i++){
number_1 += daytab[row_1][i];
}
number = number+number_1+day_1;
//第三部分为小日子到大日子之间的年的天数，进行判断是否闰年后相加
for(int i = year_2+1;i<year_1;i++){
number += NumberOfYear(i);
}
printf("%d\n",number+1);
}
}
return 0;
}
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}
};
int IsLeapYear(int year){
return (year % 4 == 0&& year %100 != 0) || (year % 400 == 0);
}
int NumberOfYear(int year){
if(IsLeapYear(year)){
return 366;
}else{
return 365;
}
}
int main(){
int date_1,date_2;
int year_1,year_2,month_1,month_2,day_1,day_2,number = 0;
int big,small;
while(scanf("%d %d",&date_1,&date_2) != EOF){
big = date_1>=date_2?date_1:date_2;
small = date_1<=date_2?date_1:date_2;
year_1 = big / 10000;
year_2 = small / 10000;
month_1 = (big - year_1*10000) / 100;
month_2 = (small - year_2*10000) / 100;
day_1 = big-year_1*10000-month_1*100;
day_2 = small-year_2*10000-month_2*100;
if(year_1 == year_2){
//判断是否闰年
int row = IsLeapYear(year_1);
for(int i = month_2;i<month_1;i++)
number+=daytab[row][i];
number = number-day_2+day_1+1;
printf("%d\n",number);
}
else{
//分为三部分
//第一部分为小日子是第多少天，并用小日子的一年天数减去小日子是第多少天
int number_1 = 0,number_2 = 0;
int row_2 = IsLeapYear(year_2);
for(int i = 1;i<month_2;i++){
number_2 += daytab[row_2][i];
}
number = NumberOfYear(year_2)-number_2-day_2;
//第二部分为大日子是第多少天，直接进行相加
int row_1 = IsLeapYear(year_1);
for(int i = 1;i<month_1;i++){
number_1 += daytab[row_1][i];
}
number = number+number_1+day_1;
//第三部分为小日子到大日子之间的年的天数，进行判断是否闰年后相加
for(int i = year_2+1;i<year_1;i++){
number += NumberOfYear(i);
}
printf("%d\n",number+1);
}
}
return 0;
}
发表于 2020-03-23 21:43:17
回复(0)
#include <iostream>
#include <algorithm>
#define leap(x) (x % 400 == 0 || (x % 4 == 0 && x % 100 != 0))
using namespace std;
int num[24][2] = {31, 31, 28, 29, 31, 31, 30, 30, 31, 31, 30, 30, 31, 31, 31, 31,
30, 30, 31, 31, 30, 30, 31, 31};//注意这里是以下标0表示一月
int cal(int y1, int m1, int d1, int y2, int m2, int d2) {
int res = 0;
while(y1 < y2 || m1 < m2 || d1 < d2) {
res++;
d1++;
if(d1 > num[m1 - 1][leap(y1)]) {
m1++;
d1 = 1;
}
if(m1 > 12) {
y1++;
m1 = 1;
}
}
return res + 1;
}
int main() {
int m, n;
while(cin >>m >>n) {
if(m > n) swap(m, n);
cout <<cal(m / 10000, (m % 10000) / 100, m % 100, n / 10000, (n % 10000) / 100, n % 100) <<endl;
}
return 0;
}
编辑于 2020-03-22 21:51:35
回复(0)
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