hdu 5455 Fang Fang（字符串处理）_fcffffcfffww-优快云博客

本文介绍了一道关于FangFang序列的编程题目，序列由特定规则生成，任务是判断给定字符串是否能由这些序列构成，并输出最少序列数量。文章提供了完整的解题思路与代码实现。

Fang Fang

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1317 Accepted Submission(s): 548

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".

题意：串可以由f,ff,cff，或者cffffffffffffffffffffffff组成，串是循环串，合法则输出最少有几串，不合法输出-1

思路：直接处理

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
char str[1000010];
int main()
{
    int T;
    scanf("%d",&T);
    for(int t=1; t<=T; t++)
    {
        scanf("%s",str);
        int len=strlen(str);
        int l=0,r=len-1,num=0;
        if(len<3)
        {
            if(len==1&&str[0]=='f')
            printf("Case #%d: 1\n",t);
            else if(len==2&&str[0]=='f'&&str[1]=='f')
            printf("Case #%d: 1\n",t);
            else printf("Case #%d: -1\n",t);
                continue;
        }
        if(str[l]=='f')
        {
            num=1;
            while(r>=0&&str[r]=='f')
                r--;
            while(l<len&&str[l]=='f')
                l++;
            if(r<0)
            {
                printf("Case #%d: %d\n",t,len%2==0?len/2:len/2+1);
                continue;
            }
            else if(r==len-1&&l==1)
            {
                printf("Case #%d: -1\n",t);
                continue;
            }
            else if(l==r&&r==len-1)
            {
                printf("Case #%d: 1\n",t);
                continue;
            }
            r--;
        }
        int flag=0;
        while(l<=r)
        {
            num++;
            if(str[l]=='c')
            {
                int q=0;
                l++;
                while(l<=r&&str[l]=='f')
                {
                    q++;
                    l++;
                }
                if(q<2)
                {
                    flag=1;
                    break;
                }
                if(l>r) break;
            }
            else
            {
                flag=1;
                break;
            }
        }
        if(flag) printf("Case #%d: -1\n",t);
        else printf("Case #%d: %d\n",t,num);
    }
    return 0;
}