CodeForce 608B Hamming Distance Sum

[题目传送门] -----------------------> http://www.codeforces.com/problemset/problem/608/B

[题目]

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|.

Sample test(s)

Input

01
00111

Output

3

Input

0011
0110

Output

2

[思路]

　　细心发现就是将A串中的每一位分别与B串中的进行比较，不同就ans++ .

　　很容易想到 两个for的暴力解法，可惜字符串长度最大都可以是200000 ， 爆搜的复杂度高达 n^2.

　　再分析发现，枚举A串的每一个字符c，只要看B串中能遇到多少不同于这个c的字符数，直接相加即可。

　　所以定义一个preOne[MAX] 数组， preOne[i] 代表B串中 从 下标 i 开始 到 sizeB － 1 的 1 出现的次数。

　　preZero[MAX]同理。

[代码]

#include <iostream>
#include <stdio.h>
#include <vector>
#include <map>
#include <string>
#include <string.h>
#include <set>
#include <algorithm>
using namespace std;
#define MAX 200010
#define LL long long 
char A[MAX];
char B[MAX];

int preOne[MAX];
int preZero[MAX];

int main()
{
    scanf("%s",A);
    scanf("%s",B);
    int sizeA = strlen(A);
    int sizeB = strlen(B);
    int sumOne = 0;
    int sumZero = 0;
    LL ans = 0;
    if(sizeA == sizeB)
    {
        for(int i = 0 ; i < sizeA ; i++)
        {
            if(A[i] != B[i]) ans++;
        }
    }
    else
    {
        for(int i = sizeB - 1 ; i >= 0 ; i--)
        {
            int temp = B[i] - '0';
            if(temp == 0)
            {
                sumZero++;
            }
            else
            {
                sumOne++;
            }
            preZero[i] = sumZero;
            preOne[i] = sumOne;
        }

        for(int i = 0 ; i < sizeA ; i++)
        {
            int temp = A[i] - '0';
            if(temp == 0)
            {
                ans += preOne[i] - preOne[sizeB - sizeA + 1 + i];
            }
            else
            {
                ans += preZero[i] - preZero[sizeB - sizeA + 1 + i];
            }
        }

    }
    cout << ans << endl;
    return 0;

}