Problem - C - Codeforces

题目描述
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| \leq |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.

输入格式
The first line contains a binary string aa ( 1 \leq |a| \leq 10^61≤∣a∣≤106 ) — the first string.

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

输出格式
Print one number — the answer.

题意翻译
给定两个由0,10,1构成的字符串aa和bb，其中|b|\leq |a|.∣b∣≤∣a∣.

对于aa的某个长度为|b|∣b∣的字串cc，记对应位不同的字符数为f(b, c).f(b,c).

如：当b=00110,c=11000b=00110,c=11000时，f(b, c) = 4,f(b,c)=4,其中第1,2,3,41,2,3,4位不同。

求所有cc中f(b,c)f(b,c)为偶数的个数。

输入输出样例
输入 #1复制

01100010
00110
输出 #1复制

3
输入 #2复制

1010111110
0110
输出 #2复制

4

题解:
考虑一次匹配的奇偶,

举个例子

1100

0011

不匹配个数是偶数

1100

1100

不匹配个数是偶数

1100

1010

不匹配个数依旧是偶数

所以位置是不用考虑的,因为涉及到变化,如果为2,相当于奇偶不变,奇数就更不用考虑了,相当于数目不变

所以可以利用异或来看一段不匹配的是不是偶数,

#include <cstdio>
#include <cstring>
#include <algorithm>
#include<iostream>
#include<vector>
#include<set>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
typedef long long ll;
#define int long long
typedef pair<int,int> PII; 
const int N = 1e6 + 10;
void solve()
{
	string a;
	string b;
	cin >> a >> b;
	int n = a.size();
	int m = b.size();
	a = " " + a;
	b = " " + b;
	int ans = 0;
	int s = 0;
	for(int i = 1;i <= m;i++)
	{
		s = s ^(a[i] - '0')^(b[i] - '0');
	}
	if(s%2 == 0)
	{
		ans++;
	}
	for(int i = m + 1;i <= n;i++)
	{
		s = s^(a[i - m] - '0')^(a[i] - '0');
		if(s%2 == 0)
		ans++;
	}
	cout << ans;
}


signed main()
{
//	ios::sync_with_stdio(0);
//	cin.tie(0);cout.tie(0);
	int t = 1;
//	cin >> t;
	while(t--)
	{
		solve(); 
	}
}

当然前缀和也可以

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=1e6+1000;
typedef long long LL;
char a[maxn],b[maxn];
LL sum[maxn];
void solve()
{
	LL alen=strlen(a+1);
	LL blen=strlen(b+1);
	for(LL i=1;i<=alen;i++){
		sum[i]=sum[i-1]+(a[i]=='1');
	}
	LL cnt1=0;
	for(LL i=1;i<=blen;i++){
		if(b[i]=='1') cnt1++;
	} 
	LL ans=0;
	for(LL i=1;i<=alen-blen+1;i++)
	{
		if ( ( (sum[i+blen-1]-sum[i-1])+cnt1)%2==0)
		{
			ans++;
		}
	}
	cout<<ans<<endl;
} 
int main(void)
{
  cin>>a+1;
  cin>>b+1;
  solve();
return 0;
}