方法思路

我们可以通过优化算法来提高效率：

1. 不需要每次都重新计算整个子串的权值，可以使用前缀和来快速计算
2. 维护两个前缀和数组：一个记录元音字母数量，一个记录辅音字母数量
3. 对于每个切割点，可以O(1)时间计算左右两部分的权值
4. 避免创建子字符串，减少内存操作

代码实现

Java

import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String s = br.readLine();
        System.out.println(countValidCuts(s));
    }
    
    public static int countValidCuts(String s) {
        int n = s.length();
        
        // 前缀和数组：vowelPrefix[i]表示前i个字符中元音的数量
        int[] vowelPrefix = new int[n + 1];
        int[] consonantPrefix = new int[n + 1];
        
        for (int i = 0; i < n; i++) {
            vowelPrefix[i + 1] = vowelPrefix[i];
            consonantPrefix[i + 1] = consonantPrefix[i];
            
            if (isVowel(s.charAt(i))) {
                vowelPrefix[i + 1]++;
            } else {
                consonantPrefix[i + 1]++;
            }
        }
        
        int count = 0;
        
        for (int i = 1; i < n; i++) {
            // 左侧子串：[0, i-1]
            int leftVowels = vowelPrefix[i];
            int leftConsonants = consonantPrefix[i];
            int leftWeight = Math.abs(leftVowels - leftConsonants);
            
            // 右侧子串：[i, n-1]
            int rightVowels = vowelPrefix[n] - vowelPrefix[i];
            int rightConsonants = consonantPrefix[n] - consonantPrefix[i];
            int rightWeight = Math.abs(rightVowels - rightConsonants);
            
            if (leftWeight == rightWeight) {
                count++;
            }
        }
        
        return count;
    }
    
    private static boolean isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
}

Python

def count_valid_cuts(s):
    n = len(s)
    
    # 前缀和数组
    vowel_prefix = [0] * (n + 1)
    consonant_prefix = [0] * (n + 1)
    
    for i in range(n):
        vowel_prefix[i + 1] = vowel_prefix[i]
        consonant_prefix[i + 1] = consonant_prefix[i]
        
        if s[i] in "aeiou":
            vowel_prefix[i + 1] += 1
        else:
            consonant_prefix[i + 1] += 1
    
    count = 0
    
    for i in range(1, n):
        # 左侧子串：[0, i-1]
        left_vowels = vowel_prefix[i]
        left_consonants = consonant_prefix[i]
        left_weight = abs(left_vowels - left_consonants)
        
        # 右侧子串：[i, n-1]
        right_vowels = vowel_prefix[n] - vowel_prefix[i]
        right_consonants = consonant_prefix[n] - consonant_prefix[i]
        right_weight = abs(right_vowels - right_consonants)
        
        if left_weight == right_weight:
            count += 1
    
    return count

s = input().strip()
print(count_valid_cuts(s))

C++

#include <iostream>
#include <string>
#include <vector>
#include <cmath>
using namespace std;

bool isVowel(char c) {
    return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}

int countValidCuts(const string& s) {
    int n = s.length();
    
    // 前缀和数组
    vector<int> vowelPrefix(n + 1, 0);
    vector<int> consonantPrefix(n + 1, 0);
    
    for (int i = 0; i < n; i++) {
        vowelPrefix[i + 1] = vowelPrefix[i];
        consonantPrefix[i + 1] = consonantPrefix[i];
        
        if (isVowel(s[i])) {
            vowelPrefix[i + 1]++;
        } else {
            consonantPrefix[i + 1]++;
        }
    }
    
    int count = 0;
    
    for (int i = 1; i < n; i++) {
        // 左侧子串：[0, i-1]
        int leftVowels = vowelPrefix[i];
        int leftConsonants = consonantPrefix[i];
        int leftWeight = abs(leftVowels - leftConsonants);
        
        // 右侧子串：[i, n-1]
        int rightVowels = vowelPrefix[n] - vowelPrefix[i];
        int rightConsonants = consonantPrefix[n] - consonantPrefix[i];
        int rightWeight = abs(rightVowels - rightConsonants);
        
        if (leftWeight == rightWeight) {
            count++;
        }
    }
    
    return count;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    string s;
    cin >> s;
    cout << countValidCuts(s) << endl;
    return 0;
}