Codeforces Round #336 (Div. 2) B. 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 is1 + 0 + 1 + 1 = 3.

The second sample case is described in the statement.

 题意 ：

输入 两个字符串a和b ，让a字符串往b里正序匹配所有与a一样长的字符然后每个字符相减求绝对值，最后绝对值求和。

我开始用for循环嵌套然后TLE成狗。

同学CW的模板，感觉挺巧妙。

所有字符串只有两个数字0和1   先用两个数值分别存放0和1出现的位置和出现的次数。

遍历a中所有字符数 ，这个数在b中对应a长度的字符串（假设为s）减去相应字符数。

假如这个数是1   在s中减去1等于0（可以不考虑） 而减去0时等于1 

那么可以用存放0的数组  用 s最后一个位置的0出现的次数减去s前一个位置0出现的次数 即为这个数所遍历的结果

0的情况反之亦然。

最后所有数相加。

  

#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;


int a[200001],b[200001],c[200001];
int m,n;
char g;


int main()
{
n=m=0;
__int64 count=0;
a[0]=b[0]=c[0]=0;
while((g=getchar())!='\n')
{
n++;
a[n]=(int)g-48;
}

while((g=getchar())!='\n')
{
m++;
b[m]=b[m-1];
c[m]=c[m-1];
if(g=='1')
b[m]++;
else c[m]++;
}

for(int i=1;i<=n;i++ )
{
if(a[i]==0)
   count+=b[m-n+i]-b[i-1];
   else
count+=c[m-n+i]-c[i-1];
}
printf("%I64d",count);
return 0;
}