Fxx and string

最新推荐文章于 2024-12-28 00:07:01 发布

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最新推荐文章于 2024-12-28 00:07:01 发布

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本文介绍了一道关于在字符串中寻找特定等比数列字符组合的问题，并提供了一个有效的解决方案。通过枚举公比和起始位置的方式，避免了暴力求解带来的效率问题。

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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：给你一个字符串，问你有多少个三元组i，j，k，满足以下条件
1.i,j,k构成等比数列
2.Si=’y’,Sj=’r’,Sk=’x’.
3.i,j,k都是整数
解题思路：我刚开始是想枚举i的位置，然后直接暴力做。但是这样做会超时，我们知道，等比数列的增长很快，所以我们可以枚举公比和i的位置。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 10;
char s[maxn];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",s);
        int len = strlen(s);
        int ans = 0;
        for(int i = 0; i < len; i++)
        {
            for(int q = 2;;q++)
            {
                int a = i + 1;
                int b = a*q;
                int c = b*q;
                if(c > len) break;
                if(s[a - 1] == 'y'&&s[b - 1] == 'r'&&s[c - 1] == 'x') ans++;
                if(s[a - 1] == 'x'&&s[b - 1] == 'r'&&s[c - 1] == 'y') ans++;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}