题解 | #字符串距离#

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

int main() {
    string s, t;
    cin >> s >> t;
    
    long long lens = s.length();
    long long lent = t.length();
    long long count_a = 0, count_b = 0, total = 0;
    int pos = 0;
    
    // 遍历字符串S
    for(int i = 0; i < lens; i++) {
        // 统计T中字符数量
        if(i < lent) {
            if(t[i] == 'a') count_a++;
            else count_b++;
        }
        
        // 计算距离
        if(s[i] == 'a') {
            total += count_b;  // 与'b'不同
        } else {
            total += count_a;  // 与'a'不同
        }
        
        // 更新滑动窗口
        if(i >= lens - lent) {
            if(t[pos++] == 'a') count_a--;
            else count_b--;
        }
    }
    
    cout << total << endl;
    return 0;
}

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        String t = sc.next();
        
        long lens = s.length();
        long lent = t.length();
        long countA = 0, countB = 0, total = 0;
        int pos = 0;
        
        // 遍历字符串S
        for(int i = 0; i < lens; i++) {
            // 统计T中字符数量
            if(i < lent) {
                if(t.charAt(i) == 'a') countA++;
                else countB++;
            }
            
            // 计算距离
            if(s.charAt(i) == 'a') {
                total += countB;  // 与'b'不同
            } else {
                total += countA;  // 与'a'不同
            }
            
            // 更新滑动窗口
            if(i >= lens - lent) {
                if(t.charAt(pos++) == 'a') countA--;
                else countB--;
            }
        }
        
        System.out.println(total);
    }
}

s = input()
t = input()

lens = len(s)
lent = len(t)
count_a = count_b = total = pos = 0

# 遍历字符串S
for i in range(lens):
    # 统计T中字符数量
    if i < lent:
        if t[i] == 'a':
            count_a += 1
        else:
            count_b += 1
    
    # 计算距离
    if s[i] == 'a':
        total += count_b  # 与'b'不同
    else:
        total += count_a  # 与'a'不同
    
    # 更新滑动窗口
    if i >= lens - lent:
        if t[pos] == 'a':
            count_a -= 1
        else:
            count_b -= 1
        pos += 1

print(total)