关于前缀和，A - Hamming Distance Sum

前缀和思想

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|.

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|.

Examples

Input

01
00111

Output

3

Input

0011
0110

Output

2

https://blog.csdn.net/chen_yuazzy/article/details/77074410

 

根本思想请看https://www.cnblogs.com/mrclr/p/8423136.html

#include<cstdio>
#include<cstring>
#define m 300000
using namespace std;
char a[m];
char b[m];
int pre0[m];
int pre1[m];
int main()
{
    int len1,len2,i;
    long long s=0;
    gets(a);
    gets(b);
    len1=strlen(a);
    len2=strlen(b);
    for(i=0;i<len2;i++)
    {
        if(b[i]=='0') {
                pre0[i]=pre0[i-1]+1;
                pre1[i]=pre1[i-1];
        }
        else if(b[i]=='1') {
                pre1[i]=pre1[i-1]+1;
                pre0[i]=pre0[i-1];
        }
    }
    for(i=0;i<len1;i++)
    {
        if(a[i]=='0')
        {
            s+=pre1[i+len2-len1]-pre1[i-1];
        }
        else if(a[i]=='1')
        {
            s+=pre0[i+len2-len1]-pre0[i-1];
        }
    }
    printf("%lld",s);
    return 0;
}