Codeforces Round #336 (Div. 2) B. Hamming Distance Sum 计算答案贡献+前缀和

B. Hamming Distance Sum

 

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：给你两个串 a,b；

对于  "0011" ， "0110"   价值就是  |0 - 0| + |0 - 1| + |1 - 1| + |1 - 0| = 0 + 1 + 0 + 1 = 2.

a的长度严格小于等于b，a从b其实对应位置开始从右移到a,b末尾位置对应，问你在这一个过程中 价值是多少

题解：我们就  计算对于b串每一个元素  所取得的价值是多少就好了，算个前缀就好

//meek///#include<bits/stdc++.h>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include<iostream>
#include<bitset>
using namespace std ;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair
typedef long long ll;

const int N = ;
const int inf = 0x3f3f3f3f;
const int MOD = ;
const double eps = 0.000001;

char a[N],b[N];
int sum[N],hou[N];
int main()
 {
     scanf("%s%s",a,b);
     int lena=strlen(a);
     for(int i=;i<lena;i++) {
         sum[i+] = sum[i]+a[i]-'';
     }
     ll ans=;
     int len=strlen(b);
     for(int i=;i<len;i++) {
        b[i]-='';
     }
     for(int i=;i<len;i++) {
          int l,r;
        if(i+>=lena) r=lena;
        else r=i+;
        if(i+>=lena) {
             if(i+lena<=len) l=;
        else l=(i+)-(len-lena);
        }
        else {
            if(len-lena>=i+) l=;
            else {
                l=i+-(len-lena);
            }
        }

        if(b[i]==) {
            ans += (r-l+)-(sum[r]-sum[l-]);
        }
        else ans+= (sum[r]-sum[l-]);
     }
     cout<<ans<<endl;
 return ;
}

代码