First Date （hnoj12952）日期计算

First Date

Time Limit: 3000ms, Special Time Limit:7500ms, Memory Limit:65536KB

Total submit users: 77, Accepted users: 46

Problem 12952 : No special judgement

Problem description

Given the last day for which the Julian calendar is in effect for some country (expressed as a Julian date), determine the next day’s Gregorian date, i.e., the first date that uses the Gregorian calendar.

Input

For each test case, the input consists of one line containing a date in the Julian calendar, formatted as YYYY-MM-DD. This date will be no earlier than October 4, 1582, and no later than October 18, 9999. The given date represents the last day that the Julian calendar is in effect for some country.

Output

For each test case, print the first Gregorian date after the calendar transition.

Sample Input

1582-10-04 1752-09-02 1900-02-25 1923-02-15

Sample Output

1582-10-15 1752-09-14 1900-03-10 1923-03-01

题意：J日历闰年只要被4整除；G日历能被4整除但不能被100整除，或者能被400整除的是闰年；

　　先在已知J的日历日期，问你G的日历显示的日期？

表示自己不会算钱，这次连日子都算不好。。。。。。。。。。。。。悲剧！

这次就被黑在这里了。。。。。无语。

题目链接：http://acm.hnu.cn/online/?action=problem&type=show&id=12952

AC代码：

#include<stdio.h>

void Print(int h,int m,int s)
{
    printf("%d",h);

    if(m>9) printf("-%d",m);
    else     printf("-0%d",m);
    if(s>9) printf("-%d\n",s);
    else     printf("-0%d\n",s);
}
int main()
{
    int i,j,d;
    int sum;
    int hh,mm,ss;
    int year,mouth,day;
    int s1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
    int s2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
while(~scanf("%d-%d-%d",&hh,&mm,&ss))
    {

        int fa,fb,fc,fd;
//           fa=hh/4;
        fb=hh/100;
        fc=hh/400;
        fd=fb-fc-1;
        year=hh;
        mouth=mm;
        day=ss+fd;
//           printf("%d\t%d\t%d\t%d\n",fa,fb,fc,fd);
        sum=0;
//           printf("%d\n",day);
        if(year%100==0&&year%400!=0&&mm<=2)
            day--;
        for(i=mm; i<13; i++)
        {
            //          sum+=s1[i];
            if(i==2&&((year%4==0&&year%100!=0)||year%400==0))
            {
                if(day>29)
                {
                    day-=29;
                    mouth++;
                    if(mouth>12)
                    {
                        mouth=1;
                        i=0;
                        year++;
                    }
                }
                else
                    break;
            }

            else if(day>s1[i])
            {
                day-=s1[i];
                mouth++;
                if(mouth>12)
                {
                    mouth=1;
                    i=0;
                    year++;
                }
            }
            else
                break;

        }
        Print(year,mouth,day);
    }
    return 0;
}

超时，和RE的代码。。。。呜呜，过不了

why？

#include<stdio.h>
#include<string.h>

#define ll __int64

int leapj(ll y)
{
    if(y%4==0)
        return 1;
    else
        return 0;
}
int leapg(ll y)
{
    if(((y%4==0)&&(y%100!=0))||(y%400==0))
        return 1;
    else
        return 0;
}

void Print(ll h,ll m,ll s)
{
    printf("%I64d",h);

    if(m>9) printf("-%I64d",m);
    else     printf("-0%I64d",m);
    if(s>9) printf("-%I64d\n",s);
    else     printf("-0%I64d\n",s);
}
int main()
{
    ll i,j,d;
    ll num;
    ll hh,mm,ss;
    ll HH,MM,SS;
    ll s1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
    ll s2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
    ll s3[13]= { 0,0, 0, 0, 0, 0, 0, 0, 0, 0,27,30,31};
    while(~scanf("%I64d-%I64d-%I64d",&hh,&mm,&ss))
    {
        ll a=(hh-1580)/4;
        if(leapj(hh)&&(mm==1||mm==2)&&ss<29)
            a--;

//        printf("a    %d\n",a);

        num=(hh-1582)*365+a+11;
        if(hh>1582)
        {
            for(j=1;j<mm; j++)
            {
                num+=s1[j];
            }
 //           if(leapj(hh)&&(mm==2&&ss==29||mm>2))
 //               num++;
            num+=ss;
        }
//        printf("num    %I64d\n",num);
//        printf("hh   %d\n",hh);
        if(hh==1582&&num<=88)
        {
            for(j=10; j<13; j++)
            {
                if(num>s3[j])
                {
                    num-=s3[j];
                    mm++;
                }
                else
                    break;
            }
            ss+=num;
            Print(hh,mm,ss);
        }

        else
        {
        //    num-=365;
//           printf("hh    %d\n",hh);
            HH=1583;

            while(num>=366)
            {
//              printf("%I64d\t%d\n",num,hh);
                num-=365;
                if(((HH%4==0)&&(HH%100!=0))||(HH%400==0))
                    num--;
                HH++;
            }

            MM=0;
            SS=0;
 //         printf("%I64d    %I64d\n",num,HH);
            if(leapg(HH))
            {
                for(j=1; j<13; j++)
                {
                    MM++;
                    if(num>s2[j])
                    {
                        num-=s2[j];
                    }
                    else
                        break;
                }
                SS=num;
            }
            else
            {
                for(j=1; j<13; j++)
                {
                    MM++;
                    if(num>s1[j])
                    {
                        num-=s1[j];
                    }
                    else
                        break;
                }
                SS=num;
            }

            Print(HH,MM,SS);
        }

    }
    return 0;
}