Codeforces #571 C Vus the Cossack and Strings

题目链接

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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|≤|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.

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Input

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

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

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Output

Print one number — the answer.

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Solution 1

这是我在考场上想出来的做法，但是写挂了

考虑串b移动之后会发生什么
例如
00110 ->
*00110
可以发现对于没移动的之前的第$i$位，如果$i-1$位和$i$位相同，那么移动后在$i$这个位置上对答案的贡献是不会变的。
那么考虑不相同的时候假设第$i$位是$0$现在匹配到$1$，$i-1$位是$1$那么移动后在$i$这个位置上对答案的贡献是$-1$。
如果第$i$位匹配到$0$，那么移动后对答案的贡献就是$+1$
如果第$i$是$1$情况类似。
也就是如果$b$串$i$位和$i-1$位不一样，每移动一次对答案就会有一个^1的贡献(答案只考虑奇偶)
对b串求一次这个

int trans = 0;
for(int i=1;i<n;i++){
	trans ^= (str[i]-'0')^(str[i+1]-'0');
}

但是$b$串移动之后，答案异或上b串的$trans$后只是更新了蓝色部分对答案的贡献。
还需要手动更新红色和绿色。

感觉我说的有点乱，可以看代码理解一下。
ps：b串的$trans$就是第一个异或最后一个，我也不知道为什么这么神奇。

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e6;
char s1[MAXN],s2[MAXN];
int n,m;
int main(){
	scanf("%s",s1+1);
	scanf("%s",s2+1);
	n=strlen(s1+1);
	m=strlen(s2+1);
	int flag = 0;
	flag ^= s2[1] ^ s2[m];
	int ans = 0;
	int tmp = 0;
	for(int i=1;i<=m;i++){
		if(s1[i] != s2[i])
			tmp ^= 1;
	}
	if(!tmp) ans ++;
	int point = m;
	while(point < n){
		point ++;
		tmp ^= flag;
		if(s1[point-m]!=s2[1])
			tmp ^= 1;
		if(s1[point] != s2[m]){
			tmp ^= 1;
		}
		if(!tmp){
			ans ++;
		}
	}
	cout<<ans;
	return 0;
}

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Solution 2

我在结束之后$Orz$了一下其他人的代码
发现了这个做法
考虑两个等长的$0/1$串$a,b$，$a$中有$s_a$个$1$,$b$中有$s_b$个$1$
那么这两个串不相同的位置的个数的就行就是$s_a+s_b$的奇偶性
为什么？
先考虑只有$0$和$1$匹配和$0$和$0$匹配，显然正确，并且不同的个数就是$s_a+s_b$
因为如果有多个$1$匹配$1$，那么不同的个数就是$s_a+s_b-k\ast 2$减掉的数是$2$的倍数，不改变奇偶性。
看完之后感觉自己是个zz

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Solution 3

这不是个fft板子题吗