[编程题]日期累加
正反两过程的综合，先将日期转换为天数，累加后再转换为日期😄
#include<iostream>
#include<cstdio>
using namespace std;
//数据预处理
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}
};
//判断平年还是闰年
bool IsLeapYear(int year){
return (year%4==0 && year%100!=0)||(year%400==0);
}
//平年闰年天数
int NumberOfYear(int year){
if(IsLeapYear(year)){
return 366;
}
else{
return 365;
}
}
//主函数
int main(){
int year,month,day;
int number;
int caseNumber;
cin>>caseNumber;
while(caseNumber--){
cin>>year>>month>>day>>number;
for(int i=0; i<month; i++){
number += daytab[IsLeapYear(year)][i];
}
number += day;//至此将加上若干天后的日期转换为数字，接下来再换回日期reverse！
month = 0;
while(number>NumberOfYear(year)){
number -= NumberOfYear(year);
year += 1;
}//对年份的处理
while(number>daytab[IsLeapYear(year)][month]){
number -= daytab[IsLeapYear(year)][month];
month += 1;
}//对月份的处理
day = number;//最后对日期的处理
printf("%04d-%02d-%02d\n",year,month,day);
}
return 0;
}
发表于 2020-02-11 11:19:23
回复(1)
import datetime m = input() time_set = [] for i in range(int(m)): temp = input().split() time_set.append(temp) for temp_time in time_set: time1 = datetime.date(int(temp_time[0]), int(temp_time[1]), int(temp_time[2])) time2 = time1 + datetime.timedelta(days = int(temp_time[3])) print(time2)
发表于 2019-03-17 15:36:46
回复(0)
#include<stdio.h>
int main()
{
int n,y,m,d,w,key;
int a[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}
};
scanf("%d",&n);
while(n--)
{
scanf("%d%d%d%d",&y,&m,&d,&w);
key=0;
if(y%4==0&&y%100!=0||y%400==0) key=1;
d=d+w;//直接把日加上能加的天数
while(d>a[key][m])//不是这个月
{
d=d-a[key][m];//减去天数
m++;//月份加1
if(m>12){//年份加一
m=1;
y++;
key=0;//重新判断闰年
if(y%4==0&&y%100!=0||y%400==0) key=1;
}
}
printf("%d-%02d-%02d\n",y,m,d);
}
}
发表于 2020-05-10 13:57:34
回复(0)
#include <cstdio>
using namespace std;
struct Date{
int y;
int m;
int d;
};
int no_month_days[12]{31,28,31,30,31,30,31,31,30,31,30,31};
int leap_month_days[12]{31,29,31,30,31,30,31,31,30,31,30,31};
bool IsLeapYear(int y){
return y%4==0&&y%100!=0||y%400==0;
}
Date operator+(Date date,int add_days){
date.d+=add_days;
while(1){
int *month_days;
month_days=IsLeapYear(date.y)?leap_month_days:no_month_days;
if(date.d<=month_days[date.m-1])
break;
else{
date.d-=month_days[date.m-1];date.m++;
if(date.m>12){
date.y++;date.m=1;
}
}
}
return date;
}
int main(){
int n,y,m,d,add_days;
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d%d%d%d",&y,&m,&d,&add_days);
Date res=Date{y,m,d}+add_days;
printf("%04d-%02d-%02d\n",res.y,res.m,res.d);
}
return 0;
}
发表于 2020-04-19 02:20:46
回复(0)
Java
import java.text.DecimalFormat;
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);
int n = scanner.nextInt();
for (int i = 0; i < n; i++) {
int year = scanner.nextInt();
int month = scanner.nextInt();
int day = scanner.nextInt();
int num = scanner.nextInt();
LocalDate date = LocalDate.of(year, month, day);
LocalDate date1 = date.plus(num, ChronoUnit.DAYS);
DecimalFormat f = new DecimalFormat("00");
System.out.println(date1.getYear()+"-"+f.format(date1.getMonth().getValue())+"-"+f.format(date1.getDayOfMonth()));
}
}
}
发表于 2020-03-19 22:57:42
回复(0)
#include<iostream>
using namespace std;
int month[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 isleap(int y)
{
if((y%4==0&&y%100!=0)||(y%400==0)) return 1; //是闰年
else return 0;
}
int main()
{
int m;
cin>>m;
for(int i=0;i<m;i++)
{
int y,m,d,n;
cin>>y>>m>>d>>n;
for(int j=0;j<n;j++)
{
int flag=isleap(y);
d++;
if(d>month[m][flag])
{
d=1;
m++;
if(m>12)
{
m=1;
y++;
}
}
}
printf("%04d-%02d-%02d\n",y,m,d);
}
return 0;
} 很好理解的一种
发表于 2020-02-13 23:28:49
回复(0)
import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Date;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Calendar calendar =Calendar.getInstance();
int m,year,month,day,days;
Scanner sc=new Scanner(System.in);
m=sc.nextInt();
String []dat =new String[m];
for(int i=0;i<m;i++) {
year=sc.nextInt();
month=sc.nextInt();
day=sc.nextInt();
days=sc.nextInt();
calendar.set(Calendar.YEAR, year);
calendar.set(Calendar.MONTH, month-1);
calendar.set(Calendar.DAY_OF_MONTH, day);
calendar.add(Calendar.DAY_OF_MONTH, days);
Date data=calendar.getTime();
DateFormat d=new SimpleDateFormat("yyyy-MM-dd");
String n=d.format(data);
dat [i]=n;
}
for(String n:dat ) {
System.out.println(n);
}
}
}
import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Date;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Calendar calendar =Calendar.getInstance();
int m,year,month,day,days;
Scanner sc=new Scanner(System.in);
m=sc.nextInt();
String []dat =new String[m];
for(int i=0;i<m;i++) {
year=sc.nextInt();
month=sc.nextInt();
day=sc.nextInt();
days=sc.nextInt();
calendar.set(Calendar.YEAR, year);
calendar.set(Calendar.MONTH, month-1);
calendar.set(Calendar.DAY_OF_MONTH, day);
calendar.add(Calendar.DAY_OF_MONTH, days);
Date data=calendar.getTime();
DateFormat d=new SimpleDateFormat("yyyy-MM-dd");
String n=d.format(data);
dat [i]=n;
}
for(String n:dat ) {
System.out.println(n);
}
}
}
发表于 2019-04-10 23:13:38
回复(1)
import java.util.Scanner;
public class Main
{
public static void main(String[] args){
Scanner input = new Scanner(System.in);
int cnt=input.nextInt();
int[] date=new int[4];
int[] date1={31,29,31,30,31,30,31,31,30,31,30,31};//闰年
int[] date2={31,28,31,30,31,30,31,31,30,31,30,31};
int[] date3=new int[3];
for(int j=0;j<cnt;j++)
{for(int i=0;i<4;i++)
date[i]=input.nextInt();
int sumday=sumday(date,date1,date2);
int count=sumday/366;
if(count<1)
countdate(sumday,date,date1,date2,date3);
else
{for(int i=0;i<count;i++)
{
if(isyears(date[0]+i))
sumday-=366;
else
sumday-=365;
}
date[0]=date[0]+count;
countdate(sumday,date,date1,date2,date3);
}
System.out.printf("%d-%02d-%02d\n", date3[0],date3[1],date3[2]);
}
input.close();
}
public static boolean isyears(int year)
{
if(year%4==0&&year%100!=0||year%400==0)
return true;
return false;
}
public static int sumday(int[] date,int[] date1,int[] date2)
{
int sumdays=0;
if(isyears(date[0]))
{
for(int i=0;i<date[1]-1;i++)
{
sumdays+=date1[i];
}
sumdays=sumdays+date[2]+date[3];
}
else {
for(int i=0;i<date[1]-1;i++)
{
sumdays+=date2[i];
}
sumdays=sumdays+date[2]+date[3];
}
return sumdays;
}
public static void countdate(int days,int[] date,int[] date1,int[] date2,int[] date3)
{
if(isyears(date[0]))
{
int j=0;
while(days-date1[j]>0)
{days-=date1[j];
j++;
}
date3[0]=date[0];
date3[1]=j+1;
date3[2]=days;
}
else
{
int j=0;
while(days-date2[j]>0)
{days-=date2[j];
j++;
}
date3[0]=date[0];
date3[1]=j+1;
date3[2]=days;
}
}
}
public class Main
{
public static void main(String[] args){
Scanner input = new Scanner(System.in);
int cnt=input.nextInt();
int[] date=new int[4];
int[] date1={31,29,31,30,31,30,31,31,30,31,30,31};//闰年
int[] date2={31,28,31,30,31,30,31,31,30,31,30,31};
int[] date3=new int[3];
for(int j=0;j<cnt;j++)
{for(int i=0;i<4;i++)
date[i]=input.nextInt();
int sumday=sumday(date,date1,date2);
int count=sumday/366;
if(count<1)
countdate(sumday,date,date1,date2,date3);
else
{for(int i=0;i<count;i++)
{
if(isyears(date[0]+i))
sumday-=366;
else
sumday-=365;
}
date[0]=date[0]+count;
countdate(sumday,date,date1,date2,date3);
}
System.out.printf("%d-%02d-%02d\n", date3[0],date3[1],date3[2]);
}
input.close();
}
public static boolean isyears(int year)
{
if(year%4==0&&year%100!=0||year%400==0)
return true;
return false;
}
public static int sumday(int[] date,int[] date1,int[] date2)
{
int sumdays=0;
if(isyears(date[0]))
{
for(int i=0;i<date[1]-1;i++)
{
sumdays+=date1[i];
}
sumdays=sumdays+date[2]+date[3];
}
else {
for(int i=0;i<date[1]-1;i++)
{
sumdays+=date2[i];
}
sumdays=sumdays+date[2]+date[3];
}
return sumdays;
}
public static void countdate(int days,int[] date,int[] date1,int[] date2,int[] date3)
{
if(isyears(date[0]))
{
int j=0;
while(days-date1[j]>0)
{days-=date1[j];
j++;
}
date3[0]=date[0];
date3[1]=j+1;
date3[2]=days;
}
else
{
int j=0;
while(days-date2[j]>0)
{days-=date2[j];
j++;
}
date3[0]=date[0];
date3[1]=j+1;
date3[2]=days;
}
}
}
发表于 2019-03-25 13:19:19
回复(0)
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <cmath>
#include <set>
#include <queue>
using namespace std;
bool pd(int y)
{
if(y % 400 == 0 || (y % 4 == 0 && y % 100 != 0))
return true;
return false;
}
int months[] ={0,31,28,31,30,31,30,31,31,30,31,30,31};
int main()
{
int n;
int nowYear,nowMonth,nowDay;
int add;
cin >> n;
for(int i = 0 ; i<n;i++)
{
cin >> nowYear >> nowMonth >> nowDay >> add;
while(add)
{
if(pd(nowYear))
months[2] = 29;
else
months[2] = 28;
if(add >= months[nowMonth] - nowDay + 1)
{
add = add - (months[nowMonth] - nowDay + 1);
nowDay = 1;
nowMonth++;
if(nowMonth > 12)
{
nowMonth = 1;
nowYear++;
}
}
else
{
nowDay += add;
add = 0;
}
}
printf("%04d-%02d-%02d\n",nowYear,nowMonth,nowDay);
}
}
发表于 2019-03-04 21:18:50
回复(0)
#include <stdio.h>
#define ISYEAP(x) x%4==0&&x%100!=0||x%400==0?1:0
int main()
{
int n,y,m,d,x;
int a[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
};
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d%d%d%d",&y,&m,&d,&x);
for(int i=0;i<x;i++)
{
d++;
if(d>a[m][ISYEAP(y)])
{
d=1;
m++;
if(m>12)
{
m=1;
y++;
}
}
}
printf("%04d-%02d-%02d\n",y,m,d);
}
return 0;
} 代码不短，但是很好理解的
发表于 2019-03-15 20:21:23
回复(4)
幼儿园代码水平献丑了
#include<iostream>
#include<cstring>
using namespace std;
bool leapyear(int y) {
if (y % 100 == 0 && y % 400 == 0 || y % 100 != 0 && y % 4 == 0)
return true;
else return false;
}
int fun(int y, int m, int d, int n) {
int sum = 0;
int mon = 0, date = 0;
int month[] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };
if (leapyear(y)) {
month[2] = 29;
}
for (int i = 0; i < m; i++) {
sum += month[i];
}
sum = sum + d + n;
while (sum >= 365) {
if (leapyear(y) && (sum / 366 != 0)) {
y++;
sum -= 366;
}
if ((!leapyear(y)) && (sum / 365 != 0)) {
y++;
sum -= 365;
}
}
if (leapyear(y)) {
month[2] = 29;
}
if (!leapyear(y)) {
month[2] = 28;
}
if (sum == 0) {
y--;
mon = 12;
date = 31;
printf("%04d-%02d-%02d\n", y, mon, date);
}
if (sum != 0) {
int temp = 0;
for (int i = 1; i <= 12; i++) {
temp += month[i];
mon++;
if (temp >= sum) {
temp -= month[i];
date = sum - temp;
break;
}
}
printf("%04d-%02d-%02d\n", y, mon, date);
}
return 0;
}
int main() {
int m, year, mon, date, n;
while (cin >> m) {
for (int i = 0; i < m; i++) {
cin >> year >> mon >> date >> n;
fun(year, mon, date, n);
}
}
return 0;
}
发表于 2020-02-16 20:40:07
回复(0)
#include <iostream>
#include <cstdio>
using namespace std;
int daytab[2][13]{//使用二维数组预先记录平年闰年天数
{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 year){//isleapyear()用于判断是否是闰年
return (((year%4==0)&&(year%100!=0))||(year%400==0));
}
int main()
{
int m,year,month,day,add;
scanf("%d", &m);
for(int i=0; i<m; i++){//读入的第一个数确定测试用例的个数
scanf("%d%d%d%d", &year, &month, &day, &add);
int days = day + add;//除去整年整月余下的天数
int row;
row = isleapyear(year);
while(days>daytab[row][month]){//从所给月份开始计数
days -= daytab[row][month++];//若余下天数大于该月的总天数，则余下天数减去该月的天数，并且月份加1
if(month==13){//当月份大于12时，年数加1，月份归一
year++;
row = isleapyear(year);//!!!年份加1，可能发生平年闰年的变化
month = 1;
}
}
printf("%04d-%02d-%02d\n", year, month, days);
}
return 0;
}
#include <cstdio>
using namespace std;
int daytab[2][13]{//使用二维数组预先记录平年闰年天数
{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 year){//isleapyear()用于判断是否是闰年
return (((year%4==0)&&(year%100!=0))||(year%400==0));
}
int main()
{
int m,year,month,day,add;
scanf("%d", &m);
for(int i=0; i<m; i++){//读入的第一个数确定测试用例的个数
scanf("%d%d%d%d", &year, &month, &day, &add);
int days = day + add;//除去整年整月余下的天数
int row;
row = isleapyear(year);
while(days>daytab[row][month]){//从所给月份开始计数
days -= daytab[row][month++];//若余下天数大于该月的总天数，则余下天数减去该月的天数，并且月份加1
if(month==13){//当月份大于12时，年数加1，月份归一
year++;
row = isleapyear(year);//!!!年份加1，可能发生平年闰年的变化
month = 1;
}
}
printf("%04d-%02d-%02d\n", year, month, days);
}
return 0;
}
发表于 2021-01-09 22:40:32
回复(0)
#include <climits>
#include <iostream>
using namespace std;
class Date
{
public:
//重写构造函数，用于cin输入的year、month、day对Date对象初始化
Date(int year=0,int month =0,int day=0)
:_year(year),
_month(month),
_day(day)
{
}
//打印输出，用于题目符合格式的输出
void Print()
{
printf("%04d-%02d-%02d\n",_year,_month,_day);
//cout<<_year<<"-"<<_month<<"-"<<_day<<endl;
}
//获取当前年月，具体有多少天，方便日期+后进位
int GetMonthDays(int year,int month)
{
int monthDays[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int retDays = monthDays[month];
if((year%4==0&&year%100!=0||year%400==0)&&month==2)
{
++retDays;
}
return retDays;
}
//运算符重载，用于Date对象进行日期+=
Date& operator+=(int x)
{
_day+=x;
while(_day>GetMonthDays(_year,_month))
{
_day-=GetMonthDays(_year,_month);
++_month;
if(_month>12)
{
_month = 1;
++_year;
}
}
return *this;
}
private:
int _year;
int _month;
int _day;
};
int main() {
int m;
cin>>m;
int icount = m;
Date d[m];
//按题目意思，先进行m组的输入，每输入一组，对一个Date对象进行初始化并进行+=
while(icount)
{
int year=0,month=0,day=0,adddays=0;
cin>>year>>month>>day>>adddays;
d[icount] = Date(year,month,day);
d[icount]+=adddays;
icount--;
}
//将m组Date对象进行打印输出
for(int i=m;i>0;--i)
{
d[i].Print();
}
}
发表于 2024-02-05 16:59:05
回复(0)
#include<stdio.h>
#include<stdbool.h>
#include<string.h>
#include<math.h>/*输入第一行表示样例个数接下来的m行中，每行4个整数，分别表示年、月、日和累加的天数。
输出：
输出m行，每行按yyyy-mm-dd的个数输出。
样例输入：
1
2008 2 3 100
样例输出：
2008-05-13*/
int IsRUN(int y){
if((y%4==0&&y%100!=0)||(y%400==0))
return 1;
else
return 0;
}
int a[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 nian(int y){
if(IsRUN(y)==1)
return 366;
else
return 365;
}
int main(){
int m,i,y,n,d,sum=0,s,m1;
scanf("%d",&n);
while(n--){
scanf("%d%d%d%d",&y,&m,&d,&s);
int row=IsRUN(y);
for(i=0;i<m;i++){
sum+=a[row][i];
}
sum=sum+d+s;
while(sum>nian(y)){
sum-=nian(y);
y++;}
m1=0;
row=IsRUN(y);
for(i=0;a[row][i]<sum;i++){
m1++;
sum-=a[row][i];
}
printf("%04d-%02d-%02d\n",y,m1,sum);
}
return 0;}
#include<stdbool.h>
#include<string.h>
#include<math.h>/*输入第一行表示样例个数接下来的m行中，每行4个整数，分别表示年、月、日和累加的天数。
输出：
输出m行，每行按yyyy-mm-dd的个数输出。
样例输入：
1
2008 2 3 100
样例输出：
2008-05-13*/
int IsRUN(int y){
if((y%4==0&&y%100!=0)||(y%400==0))
return 1;
else
return 0;
}
int a[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 nian(int y){
if(IsRUN(y)==1)
return 366;
else
return 365;
}
int main(){
int m,i,y,n,d,sum=0,s,m1;
scanf("%d",&n);
while(n--){
scanf("%d%d%d%d",&y,&m,&d,&s);
int row=IsRUN(y);
for(i=0;i<m;i++){
sum+=a[row][i];
}
sum=sum+d+s;
while(sum>nian(y)){
sum-=nian(y);
y++;}
m1=0;
row=IsRUN(y);
for(i=0;a[row][i]<sum;i++){
m1++;
sum-=a[row][i];
}
printf("%04d-%02d-%02d\n",y,m1,sum);
}
return 0;}
为什么不对折腾死我了
发表于 2024-03-24 11:55:12
回复(0)
#include <iostream>
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}
};
int isleapyear(int x) {
if (x % 400 == 0 || (x % 100 != 0 && x % 4 == 0))
return 1;
else
return 0;
}
int main() {
int n;
cin >> n;
for (int p = 0; p < n; ++p) {
int yy = 0, mm = 0, dd = 0, plus_day = 0;
scanf("%d %d %d %d", &yy, &mm, &dd, &plus_day);
int leapyear = isleapyear(yy);
int sum_day = plus_day + dd;
for (int j = mm - 1 ; j > 0 ; j--) {
sum_day += daytab[leapyear][j];
}
for (; sum_day >= 365 + isleapyear(yy) ; yy++) {
sum_day = sum_day - 365 - isleapyear(yy);
}
leapyear = isleapyear(yy);
for (int i = 1; i <= 12; ++i) {
if (daytab[leapyear][i] < sum_day) {
sum_day -= daytab[leapyear][i];
} else {
printf("%04d-%02d-%02d\n", yy, i, sum_day);
break;
}
}
}
return 0;
}
发表于 2024-03-22 17:57:33
回复(0)
#include <iostream>
using namespace std;
bool IsLeap(int y) {
return y % 4 == 0 && y % 100 != 0 || y % 400 == 0;
}
int main() {
int y,m,d,n;
int l;
cin>>l;
int month[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
while (cin >>y>>m>>d>>n) { // 注意 while 处理多个 case
int sum = 0;
n+=d;
for (int i = m; i < 13; i++) {
if (IsLeap(y)) month[2] = 29;
else month[2] = 28;
sum += month[i];
if (sum >= n) {
int d = month[i] - (sum - n);
printf("%4d-%02d-%02d\n", y, i, d);
break;
}
if(i==12){
y++;
i=0;
}
}
}
}
// 64 位输出请用 printf("%lld")
发表于 2024-03-14 11:52:57
回复(0)
我感觉比较简洁好懂
#include <bits/stdc++.h>
using namespace std;
int daytable[2][12] = {{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}, {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}} ;
bool isleap(int year) {
if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0) return true;
else return false;
}
int main() { //1967 8 17 294
int m;
cin >> m;
int year, month, day, temp;
for (int i = 0; i < m; i++) {
cin >> year >> month >> day >> temp;
day = day + temp; //311
while (day > daytable[isleap(year)][month - 1]) {
day -= daytable[isleap(year)][month - 1]; //280,250,219,189,158,127,
if (month == 12) {
month = 1; //1968,1
year++;
} else month++; //9,10,11,12,2
}
printf("%04d-%02d-%02d\n", year, month, day);
}
}
发表于 2024-03-09 22:26:46
回复(0)
#include<cstdio>
int main() {
int mday[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int m;
scanf("%d",&m);
for (int i = 0; i <m ; ++i) {
int curyear,curmonth,curday,total;
int next=0; //已经走了多少天
scanf("%d%d%d%d",&curyear,&curmonth,&curday,&total);
while (next<total){
bool isLeap=curyear%400==0||curyear%100!=0&&curyear%4==0;
if(isLeap){
mday[2]={29};
} else{
mday[2]={28};
}
curday++;
if(curday>mday[curmonth]){
curmonth++;
curday=1;
if(curmonth>12){
curyear++;
curmonth=1;
}
}
next++;
}
printf("%04d-%02d-%02d\n",curyear,curmonth,curday);
}
}
发表于 2024-03-07 21:51:32
回复(0)
//好处：不用函数重载，定义一个类，2个方法搞定，main中有2中方式，一种是全部读取了再输出，一种是读取一行输出一行，
#include <iostream>
using namespace std;
class Date {
public:
Date(int year, int month, int day) {
_year = year;
_month = month;
_day = day;
};
int getMonthDay(int year, int month)const {
if (month > 12) {
return 1;
}
int arr[13] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
if (month == 2 &&
((year % 4 == 0 && year % 100 != 0)
|| year % 400 == 0)) {
return 29;
}
return arr[month];
}
void getDate( int num)const {
Date temp(_year, _month, 1);
num = num + _day;
while (num > 0 && num > getMonthDay(temp._year, temp._month)) {
num = num - getMonthDay(temp._year, temp._month);
++temp._month;
if (temp._month == 13) {
temp._year++;
temp._month = 1;
}
}
temp._day = num;
printf("%d-%02d-%02d\n", temp._year, temp._month, temp._day);
}
private:
int _year;
int _month;
int _day;
};
//方式一
int main() {
int lineNum;
cin >> lineNum;
int *arr=new int[lineNum*4];
for (int i = 0; i < lineNum; i++) {
int a, b, c, d;
cin >> a >> b >> c >> d;
arr[i * 4 ] = a;
arr[i * 4 + 1] = b;
arr[i * 4 + 2] = c;
arr[i * 4 + 3] = d;
}
for (int i = 0; i < lineNum; i++) {
Date d1(arr[i * 4], arr[i * 4 + 1], arr[i * 4 + 2]);
d1.getDate(arr[i * 4 + 3]);
}
}
//方式二
//方式一，逐行读取并输出 2ms/308kb
int main() {
int lineNum;
cin >> lineNum;
for (int i = 0; i < lineNum; i++) {
int a, b, c, d;
cin >> a >> b >> c >> d;
Date d1(a, b, c);
d1.getDate(d);
}
// 注意 while 处理多个 case
}
int main() {
int lineNum;
cin >> lineNum;
for (int i = 0; i < lineNum; i++) {
int a, b, c, d;
cin >> a >> b >> c >> d;
Date d1(a, b, c);
d1.getDate(d);
}
// 注意 while 处理多个 case
}
发表于 2024-03-05 17:03:59
回复(0)
#include <iostream>
using namespace std;
#define IsYep(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 Year;
int Month;
int Day;
void nextDay() {
Day++;
if (Day > dayOfMonth[Month][IsYep(Year)]) {
Day = 1;
Month++;
if (Month > 12) {
Month = 1;
Year++;
}
}
}
};
int main() {
Date tmp;
int year, month, day, number;
int n;
cin >> n;
for (int j = 0; j < n; j++) {
scanf("%d%d%d%d", &year, &month, &day, &number);
tmp.Year = year;
tmp.Month = month;
tmp.Day = day;
for (int i = 0; i < number; i++) {
tmp.nextDay();
}
printf("%04d-%02d-%02d\n",tmp.Year,tmp.Month,tmp.Day);
//cout << tmp.Year << "-" << tmp.Month << "-" << tmp.Day << endl;
}
}
// 64 位输出请用 printf("%lld")
发表于 2024-02-29 16:06:02
回复(0)
问题信息
通过挑战的用户
查看代码-
2023-03-13 16:36:34
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2023-03-13 09:17:55
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2023-03-10 21:05:11
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2023-03-08 11:27:36 -
2023-03-08 10:58:16
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