codeforces 608B Hamming Distance Sum

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|.

Sample test(s)

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 is 1 + 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.

#include<bitset> //15ms
#include<map>
#include<vector>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#define inf 0x3f3f3f3f
#define mem(a,x) memset(a,x,sizeof(a))

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

inline int in()
{
int res=0;char c;
while((c=getchar())<'0' || c>'9');
while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
return res;
}
const int N=200100;

char a
,b
;
ll pre0
,pre1
;

int main()
{
gets(a+1);
gets(b+1);
int n=strlen(a+1),m=strlen(b+1);
ll ans=0;
for(int i=1;i<=m;i++)
{
if(b[i]=='1')
{
pre1[i]=pre1[i-1]+1;
pre0[i]=pre0[i-1];
}
else{
pre0[i]=pre0[i-1]+1;
pre1[i]=pre1[i-1];
}
}

for(int i=1;i<=n;i++)
{
if(a[i]=='1')
{
ans += pre0[m-n+i]-pre0[i-1];
}
else{
ans += pre1[m-n+i]-pre1[i-1];
}
}
cout<<ans;
return 0;
}