（字符串处理）Fang Fang -- hdu -- 5455 （2015 ACM/ICPC Asia Regional Shenyang Online）

链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5455

Fang Fang

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 233    Accepted Submission(s): 110

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".

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>


using namespace std;

#define N 1100000
char s[N];

int main()
{
    int t, iCase=1;
    scanf("%d", &t);

    while(t--)
    {
        int i, num=0, sum=0, flag=0, len;

        scanf("%s", s);

        len = strlen(s)-1;

        for(i=0; i<=len; i++)
        {
            if(s[i]=='f')
                num++;
            if(s[i]!='f' && s[i]!='c')
                flag = 1;
            if(s[i]=='c')
            {
                if(i==len-1 && s[0]=='f' && s[len]=='f')
                    sum ++;
                else if(i==len && s[0]=='f' && s[1]=='f')
                    sum ++;
                else if(s[i+1]=='f' && s[i+2]=='f')
                    sum++;
                else
                    flag = 1;
            }
        }

        printf("Case #%d: ", iCase++);

        if(flag)
            printf("-1
");
        else
        {
            if(sum)
                printf("%d
", sum);
            else if(len+1==num)
                printf("%d
", (num+1)/2);
            else
                printf("-1
");
        }
    }
    return 0;
}