CodeForces - 260 - BAncient Prophesy（暴力）

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
题目链接
参考题解
题目要求输出其中出现次数最多的13年到15年的日期。其中2013-12-2013可以看成20 13-12-2013，所以也出现了一个日期。

这个题目数据量1e5，直接每一位字符判断一下就可以了，看一下是否符合dd-mm-yyyy的条件，然后进一步判断年份是不是在13年到15年，判断一下月份，然后记录一下当前字符位置，这个年月出现多少次，记录年月的方法算是比较有意思。

#include <cstdio>
#include <cctype>
#include <cstring>
using namespace std;
char str[100005];
int month[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int cnt[4][16][32];

int main()
{
	int d = -1, m = -1, y = -1, id, ansd = 0;
	scanf("%s", str);
	int len = strlen(str);
	for(int i = 0; i < len - 9; i++)
    {
        //先判断是否是一个合法日期，isdigit（）函数在头文件cctype中，是处理字符串的函数集合，isdigit（）用来判断是否是一个数字
        //下面先判断是否符合格式dd-mm-yyyy不符合继续往后遍历
		if(isdigit(str[i]) && isdigit(str[i+1]) && isdigit(str[i+3]) && isdigit(str[i+4]) && isdigit(str[i+6]) && isdigit(str[i+7]) && isdigit(str[i+8]) && isdigit(str[i+9]) && str[i+2]=='-'&&str[i+5]=='-')
		{
			d=(str[i] - 48) * 10 + str[i + 1] - 48;
			m=(str[i + 3] - 48) * 10 + str[i + 4] - 48;
			y=(((str[i + 6] - 48) * 10 + str[i + 7] - 48) * 10 + str[i + 8] - 48) * 10 + str[i + 9] - 48;
			if(m > 12 || y < 2013 || y > 2015 || m < 1) continue;   //如果不是13年到15年或者是月份不对，那么跳过
			if(d > month[m] || d < 1)   continue;   //如果日期不是合法的跳过
			if(++cnt[y - 2013][m][d] > ansd) ansd = cnt[y - 2013][m][d], id = i;
		}
	}
	for(int i = 0; i < 10; i++)   putchar(str[id + i]);
	putchar('\n');
	return 0;
}