Fang Fang

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
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".

#include <cstdio>
#include <queue>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
const int oo = 1e9;
const int N = 1e6+8;
const int M = 6000;
typedef long long LL;
char str[N];
int vis[N];
int main()
{
    int T, ok, i, isc, n, first, xx=1, last;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%s", str);
        n = strlen(str);
        ok = 1;
        isc = 0;
        memset(vis, 0, sizeof(vis));
        first = last = -1;
        for(i = 0; i < n; i++)
        {
            if(str[i] != 'f' && str[i] != 'c')
            {
                ok = 0;
                break;
            }
            if(str[i] == 'c')
            {
                vis[i] = 1;
                last = i;
                if(isc == 0)
                    first = i;
                isc ++;
            }
        }
        if(ok == 0)
        {
            printf("Case #%d: -1
", xx++);
            continue;
        }

        if(isc == 0)
        {
            printf("Case #%d: %d
", xx++, (n+1)/2);
            continue;
        }

        if(isc > 0)
        {
            if(n-1-last+first < 2) ok = 0;
        }

        if(isc > 0 && ok)
        {
            for(i = first+1; i <= last; i++)
            {
                if(vis[i])
                {
                    if(i-first-1 < 2)
                        ok = 0;
                    first = i;
                }
                if(ok == 0)break;
            }
        }

        if(ok == 0) printf("Case #%d: -1
", xx++);

        else printf("Case #%d: %d
", xx++, isc);
    }
    return 0;
}