A recently found Ancient Prophesy is believed to contain the exact Apocalypse date. The prophesy is a string that only consists of digits and characters "-".
We'll say that some date is mentioned in the Prophesy if there is a substring in the Prophesy that is the date's record in the format "dd-mm-yyyy". We'll say that the number of the date's occurrences is the number of such substrings in the Prophesy. For example, the Prophesy "0012-10-2012-10-2012" mentions date 12-10-2012 twice (first time as "0012-10-2012-10-2012", second time as "0012-10-2012-10-2012").
The date of the Apocalypse is such correct date that the number of times it is mentioned in the Prophesy is strictly larger than that of any other correct date.
A date is correct if the year lies in the range from 2013 to 2015, the month is from 1 to 12, and the number of the day is strictly more than a zero and doesn't exceed the number of days in the current month. Note that a date is written in the format "dd-mm-yyyy", that means that leading zeroes may be added to the numbers of the months or days if needed. In other words, date "1-1-2013" isn't recorded in the format "dd-mm-yyyy", and date "01-01-2013" is recorded in it.
Notice, that any year between 2013 and 2015 is not a leap year.
Input
The first line contains the Prophesy: a non-empty string that only consists of digits and characters "-". The length of the Prophesy doesn't exceed 105 characters.
Output
In a single line print the date of the Apocalypse. It is guaranteed that such date exists and is unique.
Examples
Input
777-444---21-12-2013-12-2013-12-2013---444-777
Output
13-12-2013
题解：判断连续的10个字符就可以了 若符合条件 在判断 大小是否符合 在标记一下
#include<iostream>
#include<cstdio>
#include<queue>
#include<map>
#include<cstring>
#include<cctype>
using namespace std;
const int N=1e5+10;
map<int,map<int,map<int,int> > > mp;
int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
char a[N];
int dd,mm,yy,ans;
int main()
{
scanf("%s",a);
int len=strlen(a);
for(int i=0;i+9<len;i++)
{
if(isdigit(a[i])&&isdigit(a[i+1])&&isdigit(a[i+3])&&isdigit(a[i+4])
&&isdigit(a[i+6])&&isdigit(a[i+7])&&isdigit(a[i+8])&&isdigit(a[i+9])
&&a[i+2]=='-'&&a[i+5]=='-')
{
int d=(a[i]-'0')*10+a[i+1]-'0';
int m=(a[i+3]-'0')*10+a[i+4]-'0';
int y=(a[i+6]-'0')*1000+(a[i+7]-'0')*100+(a[i+8]-'0')*10+(a[i+9]-'0');
if(y>=2013&&y<=2015&&m>=1&&m<=12&&d>=1&&d<=day[m])
{
mp[y][m][d]++;
if(mp[y][m][d]>ans)
{
ans=mp[y][m][d];
yy=y,mm=m,dd=d;
}
}
}
}
printf("%02d-%02d-%04d\n",dd,mm,yy);
return 0;
}