HDU 5455 Fang Fang(2015沈阳赛区网络赛)

Fang Fang

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 685 Accepted Submission(s): 285

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 

#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
using namespace std;
const int N=1e6+10;
char s[N];
int main(){
    int t,now,i,re,c,f;
    bool ok,oc;
    scanf("%d",&t);
    for(now=1;now<=t;now++){
        scanf("%s",s);
        re=0;
        ok=1;
        oc=0;
        f=0;
        for(i=0;s[i]!='\0';i++){
            if(s[i]=='c'){       
                c=0;
                i++;
                oc=1;
                 while(s[i]=='f'){
                    c++;i++;
                }
                if(s[i]=='\0'){
                   if(c+f<2){
                       ok=0;
                       break;
                   }
                }
                else if(s[i]=='c'){
                    if(c<2){
                    ok=0;
                        break;
                    }
                }
                else{
                    ok=0;
                break;
                }
                re++;
                i--;
            }
            else if(s[i]=='f')
               f++;
            else{
                ok=0;
                break;
            }
            }    
        if(!ok)
          printf("Case #%d: -1\n",now);
        else if(!oc)
          printf("Case #%d: %d\n",now,f/2+f%2);      
        else if(ok)
           printf("Case #%d: %d\n",now,re);
    }
    return 0;
}