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+,b+))
     {
         int la=strlen(a+),lb=strlen(b+);
         res1[]=res0[]=;
         for(int i=; i<=lb; i++)
         {
             if(b[i]=='')
             {
                 res1[i]=res1[i-]+;
                 res0[i]=res0[i-];
             }
             else
             {
                 res1[i]=res1[i-];
                 res0[i]=res0[i-]+;
             }
         }
         ll ans=;
         for(int i=; i<=la; i++)
             if(a[i]=='')
                 ans+=res0[lb-(la-i)]-res0[i-];
             else
                 ans+=res1[lb-(la-i)]-res1[i-];
         printf("%I64d\n",ans);
     }
     return ;
 }