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|.

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.

直接暴力求会超时。

a字符串中的每个字符都要比较lenb - lena + 1 = cmp次，这样先算出b中0 到 cmp中1的个数存入sum。然后遍历一遍a,a[i]为1则加cmp - sum,为0则加sum,同时更新sum，即每次前进一个单位，判断b[i]和b[i + cmp]即可。

参考博客：http://www.cnblogs.com/GeekZRF/p/5130958.html 感谢！！！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

int main() {
string a, b;
cin >> a >> b;
int lena = a.size(), lenb = b.size();
int cmp = lenb - lena + 1;
long long sum = 0, ans = 0;
int i;
for(i = 0; i < cmp; i++) {
if(b[i] == '1') sum++;
}
for(i = 0; i < lena; i++) {
if(a[i] == '1') ans += cmp - sum;
else ans += sum;
if(b[i] == '1') sum--;
if(b[i + cmp] == '1') sum++;
}
printf("%I64d\n", ans);
return 0;
}