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！
 1 /******************************
 2 code by drizzle
 3 blog: www.cnblogs.com/hsd-/
 4 ^ ^    ^ ^
 5  O      O
 6 ******************************/
 7 #include<bits/stdc++.h>
 8 #include<map>
 9 #include<set>
10 #include<cmath>
11 #include<queue>
12 #include<bitset>
13 #include<math.h>
14 #include<vector>
15 #include<string>
16 #include<stdio.h>
17 #include<cstring>
18 #include<iostream>
19 #include<algorithm>
20 #pragma comment(linker, "/STACK:102400000,102400000")
21 using namespace std;
22 #define  A first
23 #define B second
24 const int mod=1000000007;
25 const int MOD1=1000000007;
26 const int MOD2=1000000009;
27 const double EPS=0.00000001;
28 typedef __int64 ll;
29 const ll MOD=1000000007;
30 const int INF=1000000010;
31 const ll MAX=1ll<<55;
32 const double eps=1e-8;
33 const double inf=~0u>>1;
34 const double pi=acos(-1.0);
35 typedef double db;
36 typedef unsigned int uint;
37 typedef unsigned long long ull;
38 int t;
39 char a[10005];
40 int main()
41 {
42     scanf("%d",&t);
43     for(int i=1;i<=t;i++)
44     {
45         int sum=0;
46         scanf("%s",a+1);
47         int len=strlen(a+1);
48         for(int j=1;j<=len;j++)
49         {
50             if(a[j]=='y')
51             {
52                 for(int k=2;;k++)
53                 {
54                     if(j*k*k>len)
55                         break;
56                     if(a[j*k]=='r'&&a[j*k*k]=='x')
57                     {
58                         sum++;
59                     }
60                 }
61             }
62             if(a[j]=='x')
63             {
64                 for(int k=2;;k++)
65                 {
66                     if(j*k*k>len)
67                         break;
68                     if(a[j*k]=='r'&&a[j*k*k]=='y')
69                     {
70                         sum++;
71                     }
72                 }
73             }
74         }
75         printf("%d\n",sum);
76     }
77     return 0;
78 }