zoukankan      html  css  js  c++  java
  • Fang Fang HDU

    Fang Fang says she wants to be remembered. 
    I promise her. We define the sequence FF of strings. 
    F0 = f",F0 = ‘‘f", 
    F1 = ff",F1 = ‘‘ff", 
    F2 = cff",F2 = ‘‘cff", 
    Fn = Fn1 + f", for n > 2Fn = Fn−1 + ‘‘f", for n > 2 
    Write down a serenade as a lowercase string SS in a circle, in a loop that never ends. 
    Spell the serenade using the minimum number of strings in FF, or nothing could be done but put her away in cold wilderness.

    InputAn positive integer TT, indicating there are TT test cases. 
    Following are TT lines, each line contains an string SS as introduced above. 
    The total length of strings for all test cases would not be larger than 106106.
    OutputThe output contains exactly TT lines. 
    For each test case, if one can not spell the serenade by using the strings in FF, output 1−1. Otherwise, output the minimum number of strings in FF to split SSaccording 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".

    题意很简单不说了。
    题解:sumf记录f的个数,然后遇见一个c判断一下是不是符合题意,如果‘c’在[0,len-1)区间内,判断一下它的下一个以及下下个是不是f,因为这个字符串是循环串,所以如果这个c是倒数第二个,判断一下最后一个和第一个字符是不是f
    如果最后一个字符是c,判断一下第一个和第二个字符是不是f,如果是,sumc++;

    不存在的时候就是sumf+sumc!=字符串的长度就行了

     代码如下:

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<cmath>
     5 #include<stack>
     6 #include<queue>
     7 #include<map>
     8 #include<algorithm>
     9 using namespace std;
    10 typedef long long ll;
    11 int main()
    12 {
    13     int T,t=1;
    14     scanf("%d",&T);
    15     while(T--)
    16     {
    17          string a;
    18          cin>>a;
    19 
    20          int len=a.length();
    21          int sumc=0,sumf=0,flag=0;
    22          for(int i = 0;i < len;i++)
    23          {
    24              if(a[i]=='f')
    25                 sumf++;
    26              else if(a[i]=='c')
    27              {
    28                  if(a[i+1]=='f'&&a[i+2]=='f')
    29                     sumc++;
    30                  if(i==len-2)
    31                  {
    32                      if(a[i+1]=='f'&&a[0]=='f')
    33                         sumc++;
    34                  }
    35                  if(i==len-1)
    36                  {
    37                      if(a[0]=='f'&&a[1]=='f')
    38                         sumc++;
    39                  }
    40              }
    41          }
    42          printf("Case #%d: ",t++);
    43          if(sumc+sumf!=len)
    44              printf("-1
    ");
    45          else
    46          {
    47              if(sumc==0) 
    48                     printf("%d
    ",sumf+1>>1); 
    49              else
    50                 printf("%d
    ",sumc);
    51          } 
    52     } 
    53     return 0;
    54 }
  • 相关阅读:
    [工作代码]dom4j解析实例
    dom4j解析xml报错:Nested exception: org.xml.sax.SAXParseException: White space is required between the processing instruction target and data.
    小米蓝牙耳机使用说明书
    猎豹浏览器(chrome内核)屏蔽视频广告
    [转载]python的常用代码模板
    [转载]存储过程
    存储过程格式及实例
    springboot(十八):解决跨域问题
    springboot(十八):CORS方式实现跨域
    springboot(十七):过滤器(Filter)和拦截器(Interceptor)
  • 原文地址:https://www.cnblogs.com/Cherry93/p/10029924.html
Copyright © 2011-2022 走看看