B. Hamming Distance Sum

题目连接：

http://www.codeforces.com/contest/608/problem/A

Description

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 si is the i-th character of s and ti 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|.

Input

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.

Output

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

Sample Input

01

00111

Sample Output

3

Hint

题意

给你一个a串，和一个b串，让A串去依次匹配b[0]-b[lena-1],b[1]-b[lena],b[2]-b[lena+].....

然后权值就是上下相减的绝对值

问你最后的权值和是多少

题解：

记录B串的前缀和，对于A串的每个字母而言，他所花费的代价，就是他移动的区间中，和他不一样的数的个数就好了

直接扫一遍就OK

代码

#include<bits/stdc++.h>

using namespace std;

#define maxn 200005

long long sum[maxn][2];

char a[maxn],b[maxn];

int A[maxn],B[maxn];

int main()

{

    scanf("%s%s",a+1,b+1);

    int len = strlen(a+1),len2 = strlen(b+1);

    for(int i=1;i<=len;i++)

        A[i]=a[i]-'0';

    for(int i=1;i<=len2;i++)

        B[i]=b[i]-'0';

    for(int i=1;i<=len2;i++)

    {

        for(int j=0;j<2;j++)

            sum[i][j]+=sum[i-1][j];

        sum[i][B[i]]++;

    }

    long long ans = 0;

    for(int i=1;i<=len;i++)

    {

        ans+=sum[len2-len+i][1-A[i]];

        ans-=sum[i-1][1-A[i]];

    }

    cout<<ans<<endl;

}