hdu 5455 Fang Fang(暴力)

Fang Fang

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 1294    Accepted Submission(s): 541

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  S  according 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

   
solution：

   
	把前两个直接拼接到最后，然后扫C的位置，看后面是否跟着两个f，注意可能含有其他字母，坑~

   


          
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1e6 + 20;
char s[maxn];
int main()
{
    int t;
    scanf("%d", &t);
    for (int k = 1; k <= t;k++)
    {
        scanf("%s", s);
        int flag = 0, len = strlen(s), ans = 0;
        s[len] = s[0]; s[len + 1] = s[1]; 
        for (int i = 0; i < len; i++)
            if (s[i] == 'c' || s[i] == 'f')continue;
            else { flag = 1; break; }
        for (int i = 0; i < len&&!flag;i++)
            if (s[i] == 'c')
            {
                if (s[i + 1] == 'f'&&s[i + 2] == 'f')
                {
                    i++;
                    while (i < len&&s[i] == 'f')
                        i++;
                    ans++; i--;
                }
                else {
                    flag = 1; 
                }
            }
        printf("Case #%d: ", k);
        if (flag == 1)printf("-1\n");
        else if (ans == 0)printf("%d\n", len / 2 + len % 2);
        else printf("%d\n", ans);
    }
    return 0;
}