題干：

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

題目大意：

有一個字符串，讓你找出其中最多的合法日期出現次數，格式為：dd-mm-yyyy。

解題報告：

? 學會了KMP，自然會接觸到BF算法吧。。不難想到這題的數據量肯定是要暴力的。于是乎又到了考驗代碼仔細度的時候了。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; char s[MAX]; char tmp[15]; string ans; int month[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31}; map<string,int> mp; map<string,int> :: iterator it; int main() {cin>>s;int len = strlen(s);for(int i = 0; i<=len-10; i++) {for(int j = 0; j<10; j++) {tmp[j] = s[j+i];}tmp[10]='\0';if(tmp[6] != '2' || tmp[7] != '0' || tmp[8] != '1') continue;if(tmp[9] != '3' && tmp[9] != '4' && tmp[9] != '5') continue;//處理年份 if(tmp[2] != '-' || tmp[5] != '-') continue;if(!isdigit(tmp[0]) || !isdigit(tmp[1]) || !isdigit(tmp[3]) || !isdigit(tmp[4]) || !isdigit(tmp[6]) || !isdigit(tmp[7]) || !isdigit(tmp[8]) || !isdigit(tmp[9])) continue;int yue = (tmp[3]-'0')*10+tmp[4]-'0';if(yue <=0 || yue > 12) continue;int day = (tmp[0]-'0')*10 + tmp[1]-'0';if(day > month[yue] || day <= 0) continue;mp[tmp]++;}it = mp.begin();int maxx = 0;for(;it!=mp.end(); ++it) {if(maxx < (it->second)) {maxx = it->second;ans= it->first;}}cout << ans << endl;return 0 ;}

總結：

? 剛開始1WA發現那個if(day > month[yue] || day <= 0) ?continue;寫成了day<0，，改過來交一發還是WA，想想可能是因為‘-’字符沒有判斷正確，于是直接把所有的都判斷一遍。，再交AC了。

總結

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