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 Fto split S according to aforementioned rules. Repetitive strings should be counted repeatedly.

Sample Input

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

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 char s[100006];
 6 int main()
 7 {
 8    int t,i,ans,j,flag,k=1,a;
 9    scanf("%d",&t);
10    getchar();
11    while (t--)
12    {
13       flag=0;
14       gets(s);
15       if (s[0]==' ')
16       {
17          printf("Case #%d: ",k);
18          k++;
19          printf("0\n");
20          continue;
21       }
22       for (i=0;s[i];i++){
23       if (s[i]!='c'&&s[i]!='f') {flag=1;break;}
24       if (s[i]=='c') break;
25       }
26       if (s[i]=='\0')
27       {
28          printf("Case #%d: ",k);
29          k++;
30          printf("%d\n",i/2+i%2);
31          continue;
32       }
33       a=i;
34       ans=1;
35       j=0;
36       for (i=i+1;s[i];i++)
37       {
38          if (s[i]!='c'&&s[i]!='f') {flag=1;break;}
39          if (s[i]!='c') j++;
40          else
41          {
42             if (j>=2)
43             {
44                j=0;
45                ans++;
46             }
47             else
48             {
49                flag=1;
50                break;
51             }
52          }
53       }
54       if (j+a<2) flag=1;
55       printf("Case #%d: ",k);
56       if (flag) printf("-1\n");
57       else printf("%d\n",ans);
58       k++;
59    }
60 }

转载于:https://www.cnblogs.com/pblr/p/4822000.html

hdu 5455 Fang Fang相关推荐

【HDU - 5455】Fang Fang（水题，有坑）
题干: Fang Fang says she wants to be remembered. I promise her. We define the sequence FF of strings. ...
HDU-5455 Fang Fang （思维）
题目链接题目:输入一个字符串,问最少能由几个Fi组成,若不能恰好由Fi组成,输出-1: 这题有一个坑点就是如果输入的字符串中有除c和f以外的其他字符时直接输出-1(刚开始做的时候没有想到这一点,被坑 ...
【 HDU - 5459】Jesus Is Here（dp）
题干: I've sent Fang Fang around 201314 text messages in almost 5 years. Why can't she make sense of w ...
2015 ACM/ICPC Asia Regional Shenyang Online题解
以下所有AC题解程序来自"仙客传奇"团队. AC题数:7/13 ABCFGJL A. Traversal AC的C++语言程序: #include <bits/stdc++. ...
递推DP HDOJ 5459 Jesus Is Here
题目传送门题意:简单来说就是sn = sn-1 + sn-2递推而来,求其中所有c字符的:∑i<j:sn[i..i+2]=sn[j..j+2]=''cff"(j−i) mod 530 ...
开启注册 | AAAI 2022论文北京预讲会，21场报告+24个Poster等你来
注册官网:https://event.baai.ac.cn/activities/228 国际人工智能会议AAAI 2022论文北京预讲会由中国中文信息学会青年工作委员会主办.智源社区和北京航空航天大 ...
蓝桥杯——机器人行走
某少年宫引进了一批机器人小车.可以接受预先输入的指令,按指令行动.小车的基本动作很简单,只有3种:左转(记为L),右转(记为R),向前走若干厘米(直接记数字). 例如,我们可以对小车输入如下的指令:1 ...
王伟storm_王伟 - 南京大学 - 化学化工学院
个人简介 2004年和2009年分别从中国科学技术大学化学系获得理学学士和分析化学博士学位.2009-2013在美国亚利桑那州立大学从事博士后研究.2013年12月起任南京大学教授.2017年获得中国 ...
PHP下拉框选择的实现方法
实现第一种PHP下拉框实现方法: < ?php //提交下拉框; //直接饱触发onchange事件的结果 $id=$_GET['myselect']; // myselect 为locati ...
基于java的学生信息管理系统(含源文件)
欢迎添加微信互相交流学习哦! 项目源码:https://gitee.com/oklongmm/biye 内容摘要 - 2 - 引言 - 4 - 学生信息管理系统开发的意义和目的 - ...

hdu 5455 Fang Fang

hdu 5455 Fang Fang相关推荐

最新文章

热门文章