Los solution to a problem Valley P2010 [date] palindrome

By: Fields

 

洛谷博 off

 

 

Knowledge Point: Analog + enumeration violence

Ideas: the title mentioned leap year
and many people believe that the leap is required to determine
in fact, contains a palindrome serial number of February 29, the top four is a leap year
then we can enumerate direct violence

Some small details:
1. The need to determine whether it exceeds 12 months
Date 2. Each month may differ
3. judgment does not require a leap year, February just need to see each year are 29 days to
4. The judgment of the month It must be 1000% / 100 to judge, or the same as in January and November, October is directly thrown away, the same as in February and December will happen (how I got 60 points). .

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<sstream>

using namespace std;

int ans;
stringstream ss;
string s;

int main()
{
    int start,end;
    cin>>start>>end;

    for(int i=start;i<=end;i++)
    {
        if(i%10000/100>12)
        {
            continue;
        }
        if(i%10000/100==1||i%10000/100==3||i%10000/100==5||i%10000/100==7||i%10000/100==8||i%10000/100==10||i%10000/100==12)
        {
            if(i%100>31)
            {
                continue;
            }
        }
        if(i%10000/100==4||i%10000/100==6||i%10000/100==9||i%10000/100==11)
        {
            if(i%100>30)
            {
                continue;
            }
        }
        if(i%10000/100==2)
        {
            if(i%100>29)
            {
                continue;
            }
        }
        if(i%10000/100==0)
        {
            continue;
        }
        if(i%100==0)
        {
            continue;
        }
        int v=i;
        int fl=0;

        for(int j=1;j<=4;j++)
        {
            v/=10;
        } 

        int x=v;
        int y=i%10000;

        int xx[5],yy[5];
        xx[1]=x%10;
        xx[2]=x%100/10;
        xx[3]=x%1000/100;
        xx[4]=x%10000/1000;

        yy[1]=y%10000/1000;
        yy[2]=y%1000/100;
        yy[3]=y%100/10;
        yy[4]=y%10;

        for(int j=1;j<=4;j++)
        {
            if(xx[j]!=yy[j])
            {
                fl=1;
                break;
            }
        }

        if(fl==0)
        {
            ans++;
        }
    }
    cost <<ans<<endl;
    return  0 ; 
}

View Code