CF608B.Hamming Distance Sum

Genos needs your help. He was asked to solve the following programming problem by Saitama:

The length of some string s is denoted |s|. The Hamming distance between two strings s and t of equal length is defined as , where s__i is the i-th character of s and t__i is the i-th character of t. For example, the Hamming distance between string "0011" and string "0110" is |0 - 0| + |0 - 1| + |1 - 1| + |1 - 0| = 0 + 1 + 0 + 1 = 2.

Given two binary strings a and b, find the sum of the Hamming distances between a and all contiguous substrings of b of length |a|.

Genos 需要你的帮助。Saitama 给他布置了如下编程题目：

字符串 $ s $ 的长度记为 $|s|$。对于两个等长的字符串 $ s $ 和 $ t $，其汉明距离定义为
，
其中 $ s_i $ 表示 $ s $ 的第 $ i $ 个字符，$ t_i $ 表示 $ t $ 的第 $ i $ 个字符。例如，字符串 "0011" 与字符串 "0110" 的汉明距离为
$ |0 - 0| + |0 - 1| + |1 - 1| + |1 - 0| = 0 + 1 + 0 + 1 = 2 $。

给定两个二进制字符串 $ a $ 和 $ b $，求 $ a $ 与 $ b $ 的所有长度为 $|a|$ 的连续子串之间的汉明距离之和。

The first line of the input contains binary string a (1 ≤ |a| ≤ 200 000).

The second line of the input contains binary string b (|a| ≤ |b| ≤ 200 000).

Both strings are guaranteed to consist of characters '0' and '1' only.

输入的第一行包含二进制字符串 $a$（$1 \le |a| \le 200\,000$）。

输入的第二行包含二进制字符串 $b$（$|a| \le |b| \le 200\,000$）。

保证两个字符串仅由字符 '0' 和 '1' 组成。

Print a single integer — the sum of Hamming distances between a and all contiguous substrings of b of length |a|.

输出一个整数——即字符串 $a$ 与字符串 $b$ 的所有长度为 $|a|$ 的连续子串之间的汉明距离之和。

For the first sample case, there are four contiguous substrings of b of length |a|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.

The second sample case is described in the statement.

对于第一个样例，字符串 b 中长度为 |a| 的连续子串共有四个：“00”、“01”、“11” 和 “11”。字符串 “01” 与 “00” 的距离为 |0 − 0| + |1 − 0| = 1；“01” 与 “01” 的距离为 |0 − 0| + |1 − 1| = 0；“01” 与 “11” 的距离为 |0 − 1| + |1 − 1| = 1。由于字符串 “11” 出现了两次，最后一项距离需计算两次。这些编辑距离之和为 1 + 0 + 1 + 1 = 3。

第二个样例已在题目描述中给出。