2017ecjtu-summer training #2 CodeForces 608B

B. Hamming Distance Sum

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

 

 

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|.

Examples

input

01
00111

output

3

input

0011
0110

output

2

Note

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 is1 + 0 + 1 + 1 = 3.

The second sample case is described in the statement.

题意: 定义两个字符串之间的距离是所有对应的每个字符相减绝对值的和, 给定a和b两个字符串, 求b中与a长度相同的所有子串与a字符串距离之和.

分析: 多写一些样例,就可以归纳出来, 其实就是a字符串中的第一个元素,分别 与b字符串中的第一个元素至第|b|-|a|+1个元素求距离的和, 再加上a字符串中的第2个元素,分别 与b字符串中的第2个元素至第|b|-|a|+2个元素求距离的和......直到a字符串中的第|a|个元素与b字符串中的第|a|个元素至第|b|个元素求距离的总和. 由于求距离很像求异或和, 因此可以先用两个数组分别保存b字符串的0个数的前缀和和1个数的前缀和, 遍历a字符串的时候 ,如果字符是1 ,那么求相应区间0的的个数,  如果字符是0 ,那么求相应区间1的的个数.

AC代码

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
#define maxn 222222
typedef long long ll;
char a[maxn],b[maxn];
int res1[maxn],res0[maxn];
int main()
{
    while(~scanf("%s%s",a+1,b+1))
    {
        int la=strlen(a+1),lb=strlen(b+1);
        res1[0]=res0[0]=0;
        for(int i=1; i<=lb; i++)
        {
            if(b[i]=='1')
            {
                res1[i]=res1[i-1]+1;
                res0[i]=res0[i-1];
            }
            else
            {
                res1[i]=res1[i-1];
                res0[i]=res0[i-1]+1;
            }
        }
        ll ans=0;
        for(int i=1; i<=la; i++)
            if(a[i]=='1')
                ans+=res0[lb-(la-i)]-res0[i-1];
            else
                ans+=res1[lb-(la-i)]-res1[i-1];
        printf("%I64d
",ans);
    }
    return 0;
}