*Fxx and string

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==数学and思维==

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杭电acm~bestcoder

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本文介绍了一个字符串匹配的问题，要求找出特定的三个字符'y', 'r', 'x'在字符串中的位置是否形成等比数列，并提供了一种有效的算法实现。

Fxx and string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 715 Accepted Submission(s): 311

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

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题意：给定一个字符串，找出符合要求的：下标成等比数列且对应字母与题中相符。注意1 2 4与4 2 1是两种

题解：一开始想到的是三重循环，不用试也知道超时，所以可以循环比例数j,i为数列最小值，随后是i*j,i*j*j

code：

#include<cstdio>
#include<cstring>
char s[10000];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
	scanf("%s",s+1);
	int l=strlen(s+1);
	int num=0;
	for(int i=1;i<=l;i++)
	{
		for(int j=2;i*j*j<=l;j++)
		{
			if(s[i]=='y'&&s[i*j]=='r'&&s[i*j*j]=='x')
			num++;
			if(s[i]=='x'&&s[i*j]=='r'&&s[i*j*j]=='y')
			num++;
		}
		}
	printf("%d\n",num);	
	} 
	return 0;
 }