题意:

Vus the Cossack has two binary strings, that is, strings that consist only of "0" and "1". We call these strings aa and bb. It is known that |b|≤|a||b|≤|a|, that is, the length of bb is at most the length of aa.

The Cossack considers every substring of length |b||b| in string aa. Let's call this substring cc. He matches the corresponding characters in bb and cc, after which he counts the number of positions where the two strings are different. We call this function f(b,c)f(b,c).

For example, let b=00110b=00110, and c=11000c=11000. In these strings, the first, second, third and fourth positions are different.

Vus the Cossack counts the number of such substrings cc such that f(b,c)f(b,c) is even.

For example, let a=01100010a=01100010 and b=00110b=00110. aa has four substrings of the length |b||b|: 0110001100, 1100011000, 1000110001, 0001000010.

  • f(00110,01100)=2f(00110,01100)=2;
  • f(00110,11000)=4f(00110,11000)=4;
  • f(00110,10001)=4f(00110,10001)=4;
  • f(00110,00010)=1f(00110,00010)=1.

Since in three substrings, f(b,c)f(b,c) is even, the answer is 33.

Vus can not find the answer for big strings. That is why he is asking you to help him.

 Input

The first line contains a binary string aa (1≤|a|≤1061≤|a|≤106) — the first string.

The second line contains a binary string bb (1≤|b|≤|a|1≤|b|≤|a|) — the second string.

 Output

Print one number — the answer.

算法分析:

主要是找到一个规律，当子串和匹配串的1的个数的奇偶性相同时，则f函数的结果就是偶数

AC代码:

import java.math.BigInteger;
import java.util.*;
public class test{
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        String str=scan.nextLine();
        String str1=scan.nextLine();
        int len=str.length();
        int len1=str1.length();
        int flag=0,num=0,num1=0;
        for(int i=0;i<len1;i++){
            if(str.charAt(i)=='1')num++;
            if(str1.charAt(i)=='1')num1++;
        }
        if(num%2==num1%2)flag++;
        for(int i=len1,j=0;i<len;i++,j++){
            if(str.charAt(i)=='1')num++;
            if(str.charAt(j)=='1')num--;
            if(num%2==num1%2)flag++;
        }
        System.out.println(flag);
    }
}