hdu 5455 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".

 

Source

2015 ACM/ICPC Asia Regional Shenyang Online

 

模拟判断，具体思路，首先他说是首尾相接的，所以先把开头第一个c之前的f移到总串的后面，这样字符串就是c开头的或全f，拍着判断即可，如果有别的字符就是-1的情况。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#define MAX 1000000
#define DMAX 1000000
#define MOD 1000000007
using namespace std;
typedef long long ll;
int n;
char s[MAX + 2],cs[MAX + 2];
int main() {
    scanf("%d",&n);
    for(int i = 1;i <= n;i ++) {
        scanf("%s",s);
        int j,len = strlen(s);
        for(j = 0;s[j];j ++) {
            if(s[j] != 'f') break;
        }
        strcpy(cs,s + j);
        s[j] = 0;
        strcat(cs,s);
        printf("Case #%d: ",i);
        if(cs[0] == 'f') printf("%d
",(len + 1) / 2);
        else {
            int ans = 0;
            for(int i = 0;cs[i];i ++) {
                if(cs[i] == 'c') {
                    if(cs[i + 1] != 'f' || cs[i + 2] != 'f') {
                        ans = -1;
                        break;
                    }
                    else {
                        ans ++;
                        i += 2;
                    }
                }
                else if(cs[i] != 'f') {
                    ans = -1;
                    break;
                }
            }
            printf("%d
",ans);
        }
    }
}