[编程题]打印日期
- 热度指数：42339 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
给出年分m和一年中的第n天，算出第n天是几月几号。
输入描述:
输入包括两个整数y(1<=y<=3000)，n(1<=n<=366)。
输出描述:
可能有多组测试数据，对于每组数据， 按 yyyy-mm-dd的格式将输入中对应的日期打印出来。
示例1
输入
2000 3 2000 31 2000 40 2000 60 2000 61 2001 60
输出
2000-01-03 2000-01-31 2000-02-09 2000-02-29 2000-03-01 2001-03-01
#include<stdio.h>//1.判断闰年2.找月数,累减n
int main()
{
int y,n,month,day,run,i;
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%d",&y,&n);
if(y%4==0&&y%100!=0||y%400==0) run=1;//判断闰年
else run=0;
for(i=1;i<=12;i++)
{
if(n>a[run][i])//天数大于整月
n-=a[run][i];//减去满月的天数
else{//不大于这个月的天数
month=i;day=n;break;}
}
printf("%d-%02d-%02d",y,month,day);//格式控制两位有效数字不够的话前补0
}
发表于 2020-03-30 17:15:47
回复(0)
建立一个数组，数组下标对应日期之差，先找到开始年份的1月1日，然后数组下标加和取得对应年月日
#include<stdio.h>
(737)#define ISLEAPYEAR(x) x%4==0 && x%100!=0 || x%400==0 ? 1 : 0
#define ABS(x) x>0 ? x : -x
int daysOfMonth[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 > daysOfMonth[month][ISLEAPYEAR(year)]) {
day = 1;
month++;
if (month > 12) {
month = 1;
year++;
}
}
}
};
Date dateCalc[2000000];
int main() {
Date temp;
temp.year = 0;
temp.month = 1;
temp.day = 1;
int cnt = 0;
while (cnt < 2000000) {
temp.nextDay();
dateCalc[cnt].year = temp.year;
dateCalc[cnt].month = temp.month;
dateCalc[cnt].day = temp.day;
cnt++;
}
int year, month=1, day=1, i_year, i_days, count=0;
while (scanf("%d%d", &i_year, &i_days) != EOF) {
int ctrl = 0;
count = i_year * 365;
while (dateCalc[count++].year != i_year);
year = dateCalc[count + i_days - 2].year;
month = dateCalc[count + i_days - 2].month;
day = dateCalc[count + i_days - 2].day;
printf("%04d-%02d-%02d", year, month, day);
}
return 0;
}
发表于 2020-03-20 02:07:55
回复(0)
Java8新的时间API确实好用
import java.text.DecimalFormat;
import java.time.LocalDate;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()){
int year = scanner.nextInt();
int days = scanner.nextInt();
LocalDate day = LocalDate.ofYearDay(year, days);
DecimalFormat f = new DecimalFormat("00");
System.out.println(day.getYear()+"-"+f.format(day.getMonth().getValue())+"-"+f.format(day.getDayOfMonth()));
}
}
}
发表于 2020-03-18 10:53:50
回复(0)
Once AC #include <bits/stdc++.h>
using namespace std;
bool isrunnian(int y) {
return (y%4==0&&y%100!=0)||(y%400==0);
}
int main() {
int y,m,d,n;
int date[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}
};
while(cin>>y>>n) {
m=1;
// d=0;
for(int i=1; i<=12; i++) {
if(n-date[isrunnian(y)][i]<=0) {
printf("%d-%02d-%02d\n",y,m,n);
break;
} else {
m++;
n-=date[isrunnian(y)][i];
}
}
}
return 0;
}
发表于 2020-03-14 20:24:06
回复(0)
#include <iostream>
#include <iomanip>
using namespace std;
bool is_rui(int year){
return year%400==0||((year%100!=0&&year%4==0));
}
int main(){
int rui[12]={31,29,31,30,31,30,31,31,30,31,30,31};
int pin[12]={31,28,31,30,31,30,31,31,30,31,30,31};
int year,month=0,day;
while(cin>>year>>day){
if(is_rui(year)){
while(day-rui[month]>0){
day-=rui[month];
++month;
}
cout<<year<<'-'<<setw(2)<<setfill('0')<<++month<<'-'<<setw(2)<<setfill('0')<<day<<endl;
}else{
while(day-pin[month]>0){
day-=pin[month];
++month;
}
cout<<year<<'-'<<setw(2)<<setfill('0')<<++month<<'-'<<setw(2)<<setfill('0')<<day<<endl;
}
}
return 0;
}
编辑于 2020-03-11 19:06:23
回复(0)
#include<stdio.h>
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 year){
return (year % 4 == 0 && year%100 != 0)||(year%400 == 0);
}
int main(){
int year,n;
while(scanf("%d %d",&year,&n)!=EOF){
int i;
for (i=0; i<13; i++ ){
if(n > month[i][isLeap(year)])
n -= month[i][isLeap(year)];
else break;
}
printf("%04d-%02d-%02d\n",year,i,n);
}
}
发表于 2020-03-09 18:21:47
回复(0)
tips：1.注意闰年判断，1900%400，不是闰年
2.1个循环解决,注意判断条件
3.注意日期格式输出，用c语言的（"%02d"，x）很方便
4.注意多组输入
#include<iostream>
#include<cstring>
using namespace std;
int main() {
int y, n;
while (cin >> y >> n) {
int sum = 0, date = 0, m = 0;
int mon[] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
if (y % 100 == 0 && y % 400 == 0 || y % 100 != 0 && y % 4 == 0)
mon[1] = 29;
for (int i = 0; i < 12; i++) {
sum += mon[i];
if (sum >= n) {
m = i + 1;
sum -= mon[i];
date = n - sum;
break;
}
}
printf("%04d-%02d-%02d\n", y, m, date);
}
}
发表于 2020-02-16 12:18:18
回复(0)
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
int year, day;
while (cin >> year >> day)
{
int d[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
if(year % 4 == 0 && year % 100 != 0 || year % 400 == 0)
d[2]++;
int month = 1;
while ((day - d[month]) > 0)
{
day -= d[month++];
}
printf("%d-%02d-%02d\n", year, month, day);
}
}
发表于 2020-02-07 18:32:30
回复(1)
#include<bits/stdc++.h>
#define isyear(x) ((x)%100!=0&&(x)%4==0||(x)%400==0)
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 main(){
int y,d,i;
while(scanf("%d %d",&y,&d)!=EOF){
for(i=0;i<13;i++)
if(d<month[i][isyear(y)]) break;
else d-=month[i][isyear(y)];
printf("%d-%02d-%02d\n",y,i,d);
}
}
发表于 2019-03-28 17:49:42
回复(0)
package com.speical.first;
import java.util.Scanner;
/**
* 给定年份和天数
*
* 给出是当年那月那日
*
* 注意闰年，提前累加好天数简单点
* 放假了，学校暖气不给力了啊，手好冷,md
* @author Special
* @time 2018/02/10 17:46:57
*/
public class Pro219 {
static int[] months = {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365};
public static boolean isLoopYear(int year) {
return year % 4 == 0 && year % 100 != 0 || year % 400 == 0;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner input = new Scanner(System.in);
while(input.hasNext()) {
int year = input.nextInt();
int days = input.nextInt();
boolean isLoop = isLoopYear(year);
int month = 1;
for(month = 1; month < 13; month++) {
if(days <= months[month] + (isLoop && month >= 2 ? 1 : 0)) {
break;
}
}
int day = days - (months[month - 1] + (isLoop && month - 1 >= 2 ? 1 : 0));
System.out.println(year + "-" + (month >= 10 ? "" : "0") + month + "-"
+ (day >= 10 ? "" : "0") + day);
}
}
}
发表于 2018-02-10 18:13:16
回复(0)
import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");
Calendar cld = Calendar.getInstance();
while (in.hasNext()) {
cld.set(in.nextInt(), 0, in.nextInt());
System.out.println(sdf.format(cld.getTime()));
}
}
}
发表于 2017-05-20 01:37:38
回复(0)
#!/usr/bin/env python
#encoding:utf-8
import sys
class test(object):
def __init__(self,year,days):
self.dict = [31,28,31,30,31,30,31,31,30,31,30,31]
self.year = year
self.days = int(days)
self.month = 0
self.day = 0
def j_run(self):
if (self.year %400 ==0) or (self.year %4 ==0 and self.year %100 !=0):
# print 'leap year'
return 1 #leap year
return 0 # nonleap year
def start(self):
if self.j_run():
self.dict[1] = 29 # Leap year has 29 days in February
# print self.dict
for a in range(12):
self.month = 0
self.day = 0
if self.days > self.dict[a]:
self.days-=self.dict[a]
else:
self.month,self.day = a+1,self.days
break
r ='%04d-%02d-%02d' % (self.year,self.month,self.day)
return r
for line in sys.stdin:
nums=line.strip().split(' ')
yy=int(nums[0])
n=int(nums[1])
a = test(yy,n)
print a.start()
发表于 2017-05-13 17:28:10
回复(1)
用数组记录每个月份下累计日数的边界值
搜索判断当前日数位于哪个区间
搜索初始位置设置为 day/30 , 这样大部分情况下只需要一次判断就能得出答案
#include <stdio.h>
int isLeapYear(int year)
{
if(year%4==0 && year%100!=0 || year%400==0)
return 1;
else
return 0;
}
int main(void)
{
int months_day_max[13] = {0,31,59,90,120,151,181,212,243,273,304,334,365};
//int months_day_max[13] = {0,31,60,91,121,152,182,213,244,274,305,335,366};
int year;
int day;
int month;
int leap;
////////////////////////////////////////////////////////////////////////
while(scanf("%d %d", &year, &day) != EOF)
{
if(day <= 31)
{
month = 1;
}
else
{
leap = isLeapYear(year);
if(leap)
months_day_max[1] = 30;
else
months_day_max[1] = 31;
int f = day / 30;
while(1)
{
int front = months_day_max[f - 1] + leap;
int current = months_day_max[f] + leap;
int back = months_day_max[f + 1] + leap;
if(day <= front)
{
f--;
}
else if(day > front && day <= current)
{
month = f;
day = day % front;
break;
}
else if(day > current && day <= back){
month = f + 1;
day = day % current;
break;
}
else
{
f++;
}
}
}
printf("%04d-%02d-%02d\n", year, month, day);
}
return 0;
}
编辑于 2017-04-20 13:32:24
回复(0)
<?php
$a = [0,31,28,31,30,31,30,31,31,30,31,30,31];//非润年
$b = [0,31,29,31,30,31,30,31,31,30,31,30,31];//闰年
while(fscanf(STDIN,"%d %d",$year,$day)){
$month = 0;
if($year%4==0&&$year%100!=0||$year%400==0){
while($day>$b[$month]){
$day -= $b[$month];
$month++;
}
printf("%04d-%02d-%02d\n",$year,$month,$day);
}else{
while($day>$a[$month]){
$day -= $a[$month];
$month++;
}
printf("%04d-%02d-%02d\n",$year,$month,$day);
}
}
发表于 2017-04-10 19:40:52
回复(0)
#include <iostream>
using namespace std;
int ping[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int run[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool isPing( int year ) {
if( year % 400 == 0 )
return false;
else if( year % 4 == 0 && year % 100 != 0 )
return false;
return true;
}
int main() {
int year, month, day, n;
while( cin >> year >> n ) {
if( isPing(year) ) {
month = 0;
while( n>0 ) {
day = n;
n -= ping[month];
month++;
}
} else {
month = 0;
while( n>0 ) {
day = n;
n -= run[month];
month++;
}
}
if( month < 10 && day >= 10 )
cout << year << "-0" << month << "-" << day << endl;
else if( month >= 10 && day < 10)
cout << year << "-" << month << "-0" << day << endl;
else if( month < 10 && day < 10 )
cout << year << "-0" << month << "-0" << day << endl;
else
cout << year << "-" << month << "-" << day << endl;
}
return 0;
}
发表于 2017-04-07 16:56:40
回复(0)
#include<iostream>
using namespace std;
bool isleap(int year)
{
if((year%4==0&&year%100!=0)||(year%400==0))
return true;
return false;
}
void getDay(int year,int num)
{
bool flag=isleap(year);
int leap[12]={31,29,31,30,31,30,31,31,30,31,30,31};
int y[12]={31,28,31,30,31,30,31,31,30,31,30,31};
int sum=0;
if(flag)
{
for(int i=0;i<12;i++)
{
sum+=leap[i];
if(sum>=num)
{
if(i+1>=10)
{
if((leap[i]-sum+num)>=10)
cout<<year<<"-"<<i+1<<"-"<<(leap[i]-sum+num)<<endl;
else
cout<<year<<"-"<<i+1<<"-0"<<(leap[i]-sum+num)<<endl;
}
else
{
if(leap[i]-sum+num>=10)
cout<<year<<"-0"<<i+1<<"-"<<(leap[i]-sum+num)<<endl;
else
cout<<year<<"-0"<<i+1<<"-0"<<(leap[i]-sum+num)<<endl;
}
break;
}
}
}
else
{
for(int i=0;i<12;i++)
{
sum+=y[i];
if(sum>=num)
{
if(i+1>=10)
{
if((y[i]-sum+num)>=10)
cout<<year<<"-"<<i+1<<"-"<<(y[i]-sum+num)<<endl;
else
cout<<year<<"-"<<i+1<<"-0"<<(y[i]-sum+num)<<endl;
}
else
{
if(y[i]-sum+num>=10)
cout<<year<<"-0"<<i+1<<"-"<<(y[i]-sum+num)<<endl;
else
cout<<year<<"-0"<<i+1<<"-0"<<(y[i]-sum+num)<<endl;
}
break;
}
}
}
}
int main()
{
int year,num;
while(cin>>year>>num)
{
getDay(year,num);
}
return 0;
}
发表于 2016-08-22 16:33:56
回复(0)
/**
只要建立一个数组，用于查表就好,把闰年的情况独立出来，也就是说把二月改一下即可。
而且天数只会最多有366天，也就是加一年而已，完全不用再考虑第二年情况！！
*/
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while(in.hasNext()){
fun(in.nextInt(), in.nextInt());
}
in.close();
}
public static void fun(int a, int b){
int[] month = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
if(a % 4 == 0 && a % 100 != 0 || a % 400 == 0)
month[1] = 29;
int i;
for(i = 0; i < 12; i++){
if(b <= month[i])
break;
else
b -= month[i];
}
if(i == 11 && b > 31)
a++;
String str = String.format("%d-%02d-%02d",a, i+1, b);
System.out.println(str);
}
}
编辑于 2017-02-06 17:48:11
回复(0)
建立两个数组分别存放闰年和非闰年的各个月份对应天数。
#include <cstdio>
using namespace std;
int a[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31},
b[13] = {0,31,29,31,30,31,30,31,31,30,31,30,31};
int main(){
int year, day, month;
while(scanf("%d%d", &year, &day) != EOF){
month=0;
if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0)
while (day > b[month])
day -= b[month++];
else
while (day > a[month]) {
day -= a[month++];
printf("%04d-%02d-%02d\n", year, month, day);
}return 0;
}
编辑于 2019-09-27 15:22:15
回复(5)
思路：
1. 先判断输入的年是闰年还是平年，闰年2月是29天，平年二月是28天
闰年的判定方法：
---能被4整除，不能被100整除 或者是 ---能被400整除
2. 定位是哪一个月
(1).定义一个天数数组，遍历天数数组，
(2).如果(n - 当前月的天数) >0
n -= 当前月的天数，跳到下一个月
直到 n<= 当前月的天数
3. 定位到月之后，再判断是哪一天，在第二步得到n的值即可
#include <iostream>
using namespace std;
void printDate(int year, int month, int day)
{
cout << year << '-' ;
if (month < 10)
{
cout << '0' << month << '-' ;
}
else cout << month << '-' ;
if (day < 10)
{
cout << '0' << day << endl;
}
else cout << day << endl;
}
int main()
{
static char 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 year, day_of_year, leap, i;
while (cin >> year >> day_of_year)
{
leap = ((year % 4 == 0) && (year % 100 != 0) || (year % 400 == 0));
for (i = 0; i < 13; i++)
{
if (day_of_year > daytab[leap][i])
{
day_of_year -= daytab[leap][i];
}
else break;
}
if (i == 13) i -= 12;
printDate(year, i, day_of_year);
}
return 0;
}
发表于 2017-05-11 18:34:04
回复(3)
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