hdu 5455__Fang Fang

Fang Fang

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 987    Accepted Submission(s): 415

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 Saccording 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

 

寫的有點醜= =

#include<cstring>
#include<cstdio>
char ch[1000000+10];
int main()
{
    int T,cas=1;
    scanf("%d%*c",&T);
    while(T--) {
        gets(ch);
        int len=strlen(ch);
        if(len==0) {
            printf("0\n");
            continue;
        }
        int ok=1,flag=0,ans=0,flag1=0,cnt=0;
        for(int i=0;i<len;i++) {
            if(ch[i]=='f'&&flag1==0) {
                cnt++;
            }
            if(ch[i]!='c'&&ch[i]!='f') {
                ok=0;
                break;
            }
            if(ch[i]=='c') {
                ans++;
                if(flag<=1&&flag1!=0) {
                    ok=0;//printf("//%d %d\n",flag,cnt);
                    break;
                }
                flag=0;flag1=1;
            }
            else if(ch[i]=='f') {
                flag++;
            }
        }
        if(flag1==0) {
            ans=(len+1)/2;
        }

        if(flag+cnt<=1) {
            ok=0;
        }
        printf("Case #%d: ",cas++);
        if(!ok) {
            printf("-1\n");
        }
        else
            printf("%d\n",ans);
    }
    return 0;
}