HDU 5455：Fang Fang 查cff个数

Fang Fang

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 945    Accepted Submission(s): 393

Problem Description

Fang Fang says she wants to be remembered.
I promise her. We define the sequence F of strings.
F0 = ‘‘f",
F1 = ‘‘ff",
F2 = ‘‘cff",
Fn = Fn−1 + ‘‘f", for n > 2
Write down a serenade as a lowercase string S in a circle, in a loop that never ends.
Spell the serenade using the minimum number of strings in F, or nothing could be done but put her away in cold wilderness.

 

Input

An positive integer T, indicating there are T test cases.
Following are T lines, each line contains an string S as introduced above.
The total length of strings for all test cases would not be larger than 106.

 

Output

The output contains exactly T lines.
For each test case, if one can not spell the serenade by using the strings in F, output −1. Otherwise, output the minimum number of strings in F to split Saccording to aforementioned rules. Repetitive strings should be counted repeatedly.

 

Sample Input

8
ffcfffcffcff
cffcfff
cffcff
cffcf
ffffcffcfff
cffcfffcffffcfffff
cff
cffc

 

Sample Output

Case #1: 3
Case #2: 2
Case #3: 2
Case #4: -1
Case #5: 2
Case #6: 4
Case #7: 1
Case #8: -1
Hint
Shift the string in the first test case, we will get the string "cffffcfffcff"
and it can be split into "cffff", "cfff" and "cff".

水题，实际上就是查cff的个数。唯一的坑就是全部是f时要判断一下。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

long long test;
char a[1000005];

int main()
{
	long long i, j, h, k, len, res, res_count;
	int flag;
	cin >> test;
	for (i = 1; i <= test; i++)
	{
		cin >> a;
		len = strlen(a);
		flag = 1;
		res = 0;

		for (h = 0; h < len; h++)
		{
			if (a[h] == 'c')
			{
				flag = 2;
				if (a[(h + 1) % len] == 'f'&&a[(h + 2) % len] == 'f')
				{
					h = h + 2;
					res++;
				}
				else
				{
					flag = 0;
					break;
				}
			}
			else if (a[h] == 'f')
			{
				continue;
			}
			else
			{
				flag = 0;
				break;
			}
		}
		if (flag == 0)
		{
			cout << "Case #" << i << ": " << -1 << endl;
		}
		else if (flag == 2)
		{
			cout << "Case #" << i << ": " << res << endl;
		}
		else
		{
			cout << "Case #" << i << ": " << (len+1)/2 << endl;
		}
	}

	return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。