【hihocoder 1477】闰秒

【题目链接】:http://hihocoder.com/problemset/problem/1477

【题意】



中文题

【题解】



首先,一年一年地加,把开始的年份和结束的年份之间的年根据是否为闰年;

加上365天的秒和366天的秒;

然后把开始的那一年的剩余天数加完;

再把结束的那一年的1月1号开始一直加到结束的年的月日;

把有闰秒的年月日;

判断一下,在不在两个日期之间;

(严格在两个日期之间,不能相等);

我的程序,算的时候,右端点严格会被算到;

在的话,ans++;



【Number Of WA】



8



【完整代码】

#include <cstdio>
#include <iostream>
#include <algorithm>
#define LL long long
using namespace std;
const int MAXN = 5000+100;
struct abc
{
    int year,month,day,hour,minute,second;
};
int n;
int a[MAXN];
abc ks,js,temp,tks,tjs;
LL rest;
int day[13];
bool rn(int x)
{
    if (!(x%4))
    {
        if (x%100)
            return true;
        else
        {
            if (x%400)
                return false;
            else
                return true;
        }
    }
    else
        return false;
}
void change(abc &ks)
{
    if (ks.second>59)
    {
        ks.second=0;
        ks.minute++;
        if (ks.minute>59)
            {
                ks.minute=0;
                ks.hour++;
                if (ks.hour == 24)
                {
                    ks.hour=0;
                    ks.day++;
                    if (ks.day > day[ks.month])
                    {
                        ks.day = 1;
                        ks.month++;
                        if (ks.month > 12)
                            {
                                ks.year++;
                                ks.month = 1;
                            }
                    }
                }
            }
    }
}
int bijiao(abc a,abc b)
{
    if (a.year != b.year)
        return a.year<b.year;
    if (a.month!=b.month)
        return a.month<b.month;
    if (a.day !=b.day)
        return a.day < b.day;
    if (a.hour!=b.hour)
        return a.hour < b.hour;
    if (a.minute!=b.minute)
        return a.minute < b.minute;
    if (a.second!=b.second)
        return a.second<b.second;
    return -1;
}
bool ok()
{
    int ju1 = bijiao(tks,temp),ju2 = bijiao(temp,tjs);
    if (ju1==1 && ju2==1)
        return true;
    return false;
}
void pd(int x)
{
    temp.year = x;
    if (ok()) rest++;
}
int main()
{
    //freopen("D:\\rush.txt","r",stdin);
    scanf("%d-%d-%d %d:%d:%d",&ks.year,&ks.month,&ks.day,&ks.hour,&ks.minute,&ks.second);
    scanf("%d-%d-%d %d:%d:%d",&js.year,&js.month,&js.day,&js.hour,&js.minute,&js.second);
    tks = ks,tjs = js;
    rest = 0;
    for (int i = ks.year+1;i <= js.year-1;i++)
        if (rn(i))
            rest+=31622400;
        else
            rest+=31536000;
    day[1] = day[3] = day[5] = day[7] = day[8] = day[10] = day[12] = 31;
    day[4] = day[6] = day[9] = day[11] = 30;
    if (ks.year<js.year)
    {
        int goal = ks.year+1;
        if (rn(ks.year))
            day[2] = 29;
        else
            day[2] = 28;
        while (ks.year<goal)
        {
            ks.second++;
            rest++;
            change(ks);
        }
        ks.year = js.year;
        if (rn(ks.year))
            day[2] = 29;
        else
            day[2] = 28;
        while (ks.month < js.month || ks.day < js.day || ks.hour < js.hour || ks.minute < js.minute)
        {
            rest++;
            ks.second++;
            change(ks);
        }
        rest+=js.second-ks.second;
    }
    else
        if (ks.year == js.year)
        {
            if (rn(ks.year))
                day[2] = 29;
            else
                day[2] = 28;
            while (ks.month < js.month || ks.day < js.day || ks.hour < js.hour || ks.minute < js.minute)
            {
                rest++;
                ks.second++;
                change(ks);
            }
            rest+=js.second-ks.second;
        }
    temp.month = 6,temp.day = 30,temp.hour = 23,temp.minute = 59,temp.second = 60;
    //6/30 23:59:60
    pd(1972);
    for (int i = 1981;i<=1983;i++) pd(i);
    pd(1985);
    for (int i = 1992;i<=1994;i++) pd(i);
    pd(1997);
    pd(2012);
    pd(2015);
    temp.month = 12,temp.day = 31;
    for (int i = 1972;i<=1979;i++) pd(i);
    pd(1987);
    for (int i = 1989;i <= 1990;i++) pd(i);
    pd(1995);
    pd(1998);
    pd(2005);
    pd(2008);
    pd(2016);
    cout << rest << endl;
    return 0;
}