POJ 2210 Metric Time【日期】

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一开始调用一个函数，结果竟然超时了，后来将闰年判断换成数组存储，就过了。可能每一次都来判断一次就比较耗时，一次都判断完就省时间了。

Description

Input

Output

Sample Input

7
0:0:0 1.1.2000
10:10:10 1.3.2001
0:12:13 1.3.2400
23:59:59 31.12.2001
0:0:1 20.7.7478
0:20:20 21.7.7478
15:54:44 2.10.20749

Sample Output

0:0:0 1.1.0
4:23:72 26.5.0
0:8:48 58.2.146
9:99:98 31.8.0
0:0:1 100.10.2000
0:14:12 1.1.2001
6:63:0 7.3.6848

Source

AC代码：

#include<stdio.h>
#include<string.h>
int leap[50010];
//int isRunNian(int year);
int DiJiTian(int year,int month,int day);

void is_leap()
{
	int i,j;
	memset(leap,0,sizeof(leap));
	for(i=2000;i<=50000;i++)
	{
		if(i%400==0||(i%4==0&&i%100!=0))
			leap[i]=1;
		else
			leap[i]=0;
	}
}

int main()
{
	int hour,minute,second,day,month,year;
	int mhour,mmin,msec,mday,mmonth,myear;
	char c1,c2,c3,c4,c5;
	int s;
	is_leap();
	scanf("%d",&s);
	while(s--)
	{
		scanf("%d%c%d%c%d%c%d%c%d%c%d",&hour,&c1,&minute,&c2,&second,&c3,&day,&c4,&month,&c5,&year);
		int i,sum=0;
		for(i=2000;i<year;i++)
		{
			if(leap[i])
				sum+=366;
			else
				sum+=365;
		}
		sum += DiJiTian(year,month,day);
		myear=sum/1000;		sum%=1000;
		mmonth=sum/100+1;	sum%=100;
		mday=sum+1;
		
		sum=(hour*3600+minute*60+second)*125/108;// 题意是说一天的时间是一样的 ，所以100000/（24*3600）=125/108 
		mhour=sum/10000;	sum%=10000;
		mmin=sum/100;	sum%=100;
		msec=sum;
		printf("%d%c%d%c%d%c%d%c%d%c%d\n",mhour,c1,mmin,c2,msec,c3,mday,c4,mmonth,c5,myear);
	}
	return 0;
} 


//int isRunNian(int year)
//{
//	if(year%400==0||(year%4==0&&year%100!=0))
//		return 1;
//	else
//		return 0;
//}

int DiJiTian(int year,int month,int day)//求当前这已经过了多少天
{
	int i,sum=0;
	for(i=1;i<month;i++)
	{
		if(i==1||i==3||i==5||i==7||i==8||i==10||i==12)
			sum+=31;
		else if(i==4||i==6||i==9||i==11)
			sum+=30;
		else if(i==2)
		{
			if(leap[year])
				sum+=29;
			else
				sum+=28;
		}
	}
	sum+=day-1;
	return sum;
}

只是调用了一个函数，就超时，没啥

超时代码：

#include<stdio.h>
int isRunNian(int year);
int DiJiTian(int year,int month,int day);
int main()
{
	int hour,minute,second,day,month,year;
	int mhour,mmin,msec,mday,mmonth,myear;
	char c1,c2,c3,c4,c5;
	int s;
	scanf("%d",&s);
	while(s--)
	{
		scanf("%d%c%d%c%d%c%d%c%d%c%d",&hour,&c1,&minute,&c2,&second,&c3,&day,&c4,&month,&c5,&year);
		int i,sum=0;
		for(i=2000;i<year;i++)
		{
			if(isRunNian(i))
				sum+=366;
			else
				sum+=365;
		}
		sum += DiJiTian(year,month,day);
		myear=sum/1000;		sum%=1000;
		mmonth=sum/100+1;	sum%=100;
		mday=sum+1;
		
		sum=(hour*3600+minute*60+second)*125/108;// 题意是说一天的时间是一样的 ，所以100000/（24*3600）=125/108 
		mhour=sum/10000;	sum%=10000;
		mmin=sum/100;	sum%=100;
		msec=sum;
		printf("%d%c%d%c%d%c%d%c%d%c%d\n",mhour,c1,mmin,c2,msec,c3,mday,c4,mmonth,c5,myear);
	}
	return 0;
} 


int isRunNian(int year)
{
	if(year%400==0||(year%4==0&&year%100!=0))
		return 1;
	else
		return 0;
}

int DiJiTian(int year,int month,int day)//求当前这已经过了多少天
{
	int i,sum=0;
	for(i=1;i<month;i++)
	{
		if(i==1||i==3||i==5||i==7||i==8||i==10||i==12)
			sum+=31;
		else if(i==4||i==6||i==9||i==11)
			sum+=30;
		else if(i==2)
		{
			if(isRunNian(year))
				sum+=29;
			else
				sum+=28;
		}
	}
	sum+=day-1;
	return sum;
}

POJ 2210 Metric Time【日期】

标签：

原文地址：http://blog.csdn.net/qq_16767427/article/details/44947431