HDU 5944 Fxx and string（暴力/枚举）-优快云博客

本文介绍了一个关于字符串匹配的问题，要求找出符合条件的三元组数量。通过枚举等比数列中的公比来优化算法效率。

Fxx and string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1007 Accepted Submission(s): 422

Description

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

思路

题意：青年理论计算机科学家Fxx得到了一个只包含小写字母的字符串。字符串的长度为

题解：直接暴力枚举，重点在于选择合适的枚举量，如果去暴力字符串的话，肯定T，但是枚举公比的话，效率大大的提高，因为其增长速度很快

#include<bits/stdc++.h>
using namespace std;
const int maxn = 10005;
char str[maxn];

int main()
{
	//freopen("input.txt","r",stdin);
	int T;
	scanf("%d",&T);
	while (T--)
	{
		scanf("%s",str+1);
		int len = strlen(str+1);
		int res = 0;
		for (int i = 1;i <= len && i <= len/4;i++)
		{
			for (int j = 2;i*j*j <= len;j++)
			{
				if (str[i] == 'y' && str[i*j] == 'r' && str[i*j*j] == 'x')	res++;
				if (str[i] == 'x' && str[i*j] == 'r' && str[i*j*j] == 'y')	res++;
			}
		}
		printf("%d\n",res);
	}
	return 0;
}

转载于:https://www.cnblogs.com/ZhaoxiCheung/p/6123517.html