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;
}