hdoj 5455 Fang Fang 【字符串简单题】

最新推荐文章于 2022-03-28 14:06:27 发布

原创最新推荐文章于 2022-03-28 14:06:27 发布 · 541 阅读

0 ·

CC 4.0 BY-SA版权

字符串简单题专栏收录该内容

34 篇文章

订阅专栏

本文深入探讨了AI音视频处理领域中的关键技术，特别是视频分割与语义识别，详细解释了如何利用这些技术实现更智能的音视频应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Fang Fang

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 403 Accepted Submission(s): 175

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".

我也是醉了，提交3次过，CE2次。杭电O__O "…

字符串F[0] = f，F[1] = ff，F[2] = cff，F[n] = F[n-1] + f。

题意：给你一个成环的字符串str，现在问你串str能否由F串组成，若不可以输出-1，反之输出组成str串需要的最少的F串。

没有坑的题都简单。。。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 1000000+10
using namespace std;
char str[MAXN];
int main()
{
    int t, k = 1;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%s", str);
        int len = strlen(str);
        bool flag = true;
        int pos = -1;
        for(int i = 0; i < len; i++)
        {
            if(str[i] != 'c' && str[i] != 'f')//其它字符出现
            {
                flag = false;
                break;
            }
            if(pos != -1)
                continue;
            if(str[i] == 'c')
                pos = i;//第一个c出现的位置
        }
        printf("Case #%d: ", k++);
        if(!flag)//其它字符出现
        {
            printf("-1\n");
            continue;
        }
        int num;//组成串的最少个数
        if(pos == -1)//c没有出现
        {
            num = len / 2;
            if(len & 1)
                num++;
            printf("%d\n", num);
            continue;
        }
        for(int i = 0; i < pos; i++)
            str[len++] = 'f';
        str[len] = '\n';//把字符f放在后面
        num = 0;
        for(int i = pos; i < len;)
        {
            if(!flag)//不能组成
                break;
            if(str[i] == 'c')
            {
                int sum = 0;//记录c后面f的个数
                i++;
                while(str[i] == 'f' && i < len)
                {
                    sum++;
                    i++;
                }
                if(sum <= 1)
                {
                    flag = false;
                    break;
                }
                num++;
            }
        }
        if(flag)
            printf("%d\n", num);
        else
            printf("-1\n");
    }
    return 0;
}