Contest100000578-《算法笔记》3.4小节——入门模拟->日期处理Contest100000578 《算法笔记》3.4小节——入门模拟->日期处理

文章目录

• Contest100000578 《算法笔记》3.4小节——入门模拟->日期处理
• • 1928-ProblemA-日期差值
  • 1929-ProblemB-Day of Week
  • 1931-ProblemC-打印日期
  • 2026-ProblemD-日期类
  • 2063-ProblemE-日期累加

Contest100000578 《算法笔记》3.4小节——入门模拟->日期处理

1928-ProblemA-日期差值

来自 http://codeup.cn/contest.php?cid=100000578

题析：经典的日期模拟，注意平年闰年的差别，两个时间点之间的差值用较小的时间不断累加来追上较大的时间，累加时注意到达界限时上一个级别的更新。

//1928ProblemA日期差值 
#include <iostream>
#include <stdlib.h>
#include <cstring>
using namespace std;

int monthDay[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}
		};
bool isLeap(int year)
{
	if((year%4==0&&year%100!=0) || year%400==0)
	{
		return true;
	}
	return false;
}
int main()
{
	int time1,year1,month1,day1;
	int time2,year2,month2,day2;
	while(scanf("%d%d",&time1, &time2) != EOF)
	{
		if(time1>time2)
		{
			int temp=time1;
			time1=time2;
			time2=temp;
		}
		year1 = time1/10000;
		month1 = time1%10000/100;
		day1 = time1%100;
		
		year2 = time2/10000;
		month2 = time2%10000/100;
		day2 = time2%100;
		int count=0;//计数值 
		while(year1<year2 || month1<month2 || day1<day2)//较小的时间点累加追较大的时间点
		{
			day1++;
			if(day1 > monthDay[month1][isLeap(year1)])//天数满月加一
			{
				month1++;
				day1=1;		
			}			
			if(month1 > 12)//月数满年加一
			{
				year1++;
				month1=1;
			}
			count++;
		}
		count=count+1;//如果两个日期是连续的我们规定他们之间的天数为两天
		printf("%d\n",count);
	}
	return 0;
}


           

1929-ProblemB-Day of Week

来自 http://codeup.cn/contest.php?cid=100000578

题析：主要涉及基姆拉尔森计算公式，另外还需要注意：
字符数组的初始化问题
字符串比较函数strcmp==0时表示相等


//1929ProblemBDay of Week
#include <iostream>
#include <cstdlib>
#include <cstring>
using namespace std;
char monthName[13][20] = {{},
	{"January"},{"February"},{"March"},{"April"},{"May"},
	{"June"},{"July"},{"August"},{"September"},{"October"},
	{"November"},{"December"}
};//月份 

char weekday[7][20]={{"Monday"},{"Tuesday"},{"Wednesday"},{"Thursday"},{"Friday"},
					{"Saturday"},{"Sunday"}
				};//星期 
/*
//基姆拉尔森计算公式外文名是Kim larsson calculation formula。
// W= (d+2*m+3*(m+1)/5+y+y/4-y/100+y/400+1)%7 //C++计算公式
在公式中d表示日期中的日数，m表示月份数，y表示年数。
注意：在公式中有个与其他公式不同的地方：
把一月和二月看成是上一年的十三月和十四月，
例：如果是2004-1-10则换算成：2003-13-10来代入公式计算。
*/
int main()
{
	int W,d,m,y;
	int day;
	char month[15];
	int year;
	while(scanf("%d%s%d",&day, month, &year) != EOF)
	{
		d=day;
		y=year;
		for(int i=1;i<=12;i++)
		{
			if(strcmp(monthName[i],month)==0)//strcmp字符串相等时值为0 
			{
				m=i;
			}
		}
		if(m==1 || m==2)//注意此处 把一月和二月看成是上一年的十三月和十四月，
		{
			m+=12;
			y--;
		}
		W=(d+2*m+3*(m+1)/5+y+y/4-y/100+y/400)%7;
		printf("%s\n",weekday[W]);
	}
	return 0;
}


           

1931-ProblemC-打印日期

来自 http://codeup.cn/contest.php?cid=100000578

题析：日期问题的变形，需注意：
Num--与day++的同步，此处不会出现year++的情况
打印日期的格式，需scanf的格式化输出
但法一为什么50%错误率还需进一步探讨

//1931ProblemC打印日期 
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int monthDay[13][10]={
	{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}
};
bool isLeap(int year)
{
	return (year%4==0 && year%100!=0 || year%400==0);
}
int main()
{
	int year;
	int num;
	while(scanf("%d%d",&year, &num) != EOF)//输入参数 
	{
		int day=0;//天数计数 
		int month=1;//月份计数 
		//法二 
		while(num>0)
		{
			num--;
			day++;
			if(day > monthDay[month][isLeap(year)])
			{
				day=1;
				month++; 
			}
		}
		printf("%04d-%02d-%02d\n",year,month,day);
		/*
		//法一 
		for(i=1;i<12;i++)
		{
			if(day <= num)
				day += monthDay[i][isLeap(year)];
			else
				break;
		}
		if(day<num)
		{
			day=num-day;
			printf("%04d-%02d-%02d\n",year,12,day);
		}
		else
		{
			i--;
			day=num + monthDay[i][isLeap(year)]-day;
			//打印输出年月日 
			printf("%04d-%02d-%02d\n",year,i,day);
		}
		*/
		
	}
	
	return 0;
}

           

2026-ProblemD-日期类

来自 http://codeup.cn/contest.php?cid=100000578

题析：简单的日期类问题，需注意：
多点测试n--
输出格式%04%02%02

//2026ProblemD日期类 
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int monthDay[13][10]={
			{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}
};
bool isLeap(int year)
{
	return (year%4==0&&year%100!=0 || year%400==0);
}

int main()
{
	int n;
	scanf("%d",&n);
	int year,month,day;
	while(n--)
	{
		scanf("%d%d%d",&year, &month,&day);
		day++;
		if(day>monthDay[month][isLeap(year)])
		{
			day=1;
			month++;
		}
		if(month>12)
		{
			month=1;
			year++;
		}
		printf("%04d-%02d-%02d\n",year,month,day);
	}
	return 0;
}


           

2063-ProblemE-日期累加

来自 http://codeup.cn/contest.php?cid=100000578

题析：
多点测试，n--
Num--与day++（涉及进位month++和year++）同步

//2063ProblemE日期累加
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int monthDay[13][10]={{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}
};
bool isLeap(int year)
{
	return (year%4==0&&year%100!=0 || year%400==0);
}

int main()
{
	int year,month,day,num;
	int n;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d%d%d%d",&year,&month,&day,&num);
		while(num--)
		{
			day++;
			if(day>monthDay[month][isLeap(year)])
			{
				day=1;
				month++;
			}
			if(month>12)
			{
				month=1;
				year++;
			}
		}
		printf("%04d-%02d-%02d\n",year,month,day);
	}
	return 0;
}