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=1000000007;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef __int64 ll;
const ll MOD=1000000007;
const int INF=1000000010;
const ll MAX=1ll<<55;
const double eps=1e-8;
const double inf=~0u>>1;
const double pi=acos(-1.0);
typedef double db;
typedef unsigned int uint;
typedef unsigned long long ull;
int t;
char a[10005];
int main()
{
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        int sum=0;
        scanf("%s",a+1);
        int len=strlen(a+1);
        for(int j=1;j<=len;j++)
        {
            if(a[j]=='y')
            {
                for(int k=2;;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=2;;k++)
                {
                    if(j*k*k>len)
                        break;
                    if(a[j*k]=='r'&&a[j*k*k]=='y')
                    {
                        sum++;
                    }
                }
            }
        }
        printf("%d
",sum);
    }
    return 0;
}