HDU 5944 暴力

Fxx and string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 507    Accepted Submission(s): 224

Problem Description

Young theoretical computer scientist Fxx get a string which contains lowercase letters only.

The string S

contains n

lowercase letters S1S2…Sn

.Now Fxx wants to know how many three tuple (i,j,k)

there are which can meet the following conditions:

1、i,j,k

are adjacent into a geometric sequence.

2、Si=

'y

',Sj=

'r

',Sk=

'x

'.

3.Either j|i or j|k

 

Input

In the first line, there is an integer T(1≤T≤100)

indicating the number of test cases.

T

lines follow, each line contains a string, which contains only lowercase letters.(The length of string will not exceed 10000

).

 

Output

For each case, output the answer.

 

Sample Input

2

xyyrxx

yyrrxxxxx

 

Sample Output

0

2

 

Source

BestCoder Round #89

 

题意：

题解：给你一个字符串暴力求 有多少组 ‘y’ ‘r’ ‘x’ 位置成等比数列  注意逆序！

 /******************************
 code by drizzle
 blog: www.cnblogs.com/hsd-/
 ^ ^    ^ ^
  O      O
 ******************************/
 #include<bits/stdc++.h>
 #include<map>
 #include<set>
 #include<cmath>
 #include<queue>
 #include<bitset>
 #include<math.h>
 #include<vector>
 #include<string>
 #include<stdio.h>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #pragma comment(linker, "/STACK:102400000,102400000")
 using namespace std;
 #define  A first
 #define B second
 const int mod=;
 const int MOD1=;
 const int MOD2=;
 const double EPS=0.00000001;
 typedef __int64 ll;
 const ll MOD=;
 const int INF=;
 const ll MAX=1ll<<;
 const double eps=1e-;
 const double inf=~0u>>;
 const double pi=acos(-1.0);
 typedef double db;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 int t;
 char a[];
 int main()
 {
     scanf("%d",&t);
     for(int i=;i<=t;i++)
     {
         int sum=;
         scanf("%s",a+);
         int len=strlen(a+);
         for(int j=;j<=len;j++)
         {
             if(a[j]=='y')
             {
                 for(int k=;;k++)
                 {
                     if(j*k*k>len)
                         break;
                     if(a[j*k]=='r'&&a[j*k*k]=='x')
                     {
                         sum++;
                     }
                 }
             }
             if(a[j]=='x')
             {
                 for(int k=;;k++)
                 {
                     if(j*k*k>len)
                         break;
                     if(a[j*k]=='r'&&a[j*k*k]=='y')
                     {
                         sum++;
                     }
                 }
             }
         }
         printf("%d\n",sum);
     }
     return ;
 }