C. Vus the Cossack and Strings

 

 题面：

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

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

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

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

For example, let $$$a = 01100010$$$ and $$$b = 00110$$$. $$$a$$$ has four substrings of the length $$$|b|$$$: $$$01100$$$, $$$11000$$$, $$$10001$$$, $$$00010$$$.

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

Since in three substrings, $$$f(b, c)$$$ is even, the answer is $$$3$$$.

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

 /******** 

 http://codeforces.com/contest/1186/problem/C

 题意：给2个0，1串，a串比b串长。
 问a串有多少个和b串等长且不同的字符个数为偶数的串。
 思路：首先比较出a中第一个串（记为a0）和b的不同字符个数（记为ans0）。然后a中第二个串（记为a1）将不在与b比较而是与a中第一个串比较。
 这里分2种情况：    （1）：a1[i]==a0[i]时不用管；（因为此情况时，无论a0[i]与b[i]是否相同，a1[i]与b[i]的情况都与前者一样）
                 （2）：a1[i]!=a0[i]时记录这种情况个数(记为s1)。（a1[i]与b[i]的情况恰好与a0[i]与b[i]的情况相反）
                 如果(ans0+s1)%2==0该串就是答案之一；
 那么对于任意一个a的与b大小相等的串到底是不是答案，我们都可以通过（前一个串的不同字符个数+该串中（2）情况的个数）来判断。
 而对于任意一个串， （2）情况的个数，都可以通过预处理0（1）出来。                
 ****/
 
 
 #include<bits/stdc++.h>

 using namespace std;
 int s[1000009],ans[1000009];
int main()
{
      string sa,sb;
      int la,lb;
     cin>>sa>>sb;
     la=sa.size( ),lb=sb.size();
     s[0]=0;
      for(int i=1;i<la;i++)
      {
          if(sa[i]==sa[i-1])s[i]=s[i-1];
          else s[i]=s[i-1]+1;
      }
      int k=0;
      for(int i=0;i<lb;i++)
      {
         if(sa[i]!=sb[i])ans[lb-1]++;
      }
      if(ans[lb-1]%2==0)ans[lb-1]=0;
      for(int i=lb;i<la;i++)
      {
          ans[i]=(ans[i-1]+s[i]-s[i-lb])%2;
      }
      int tans=0;
      for(int i=lb-1;i<la;i++)
      {
          if(ans[i]==0) tans++;
      }
      cout<<tans<<endl;
     return 0;
 }