【Codeforces Round 336 (Div 2) B】【水题 前缀和】Hamming Distance Sum 双字符串所有位置匹配下的差异数之和

B. Hamming Distance Sum

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}
template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}
const int N=2e5+10,M=0,Z=1e9+7,ms63=0x3f3f3f3f;
int casenum,casei;
char a
,b
;
int f[2]
;
int main()
{
while(~scanf("%s%s",a+1,b+1))
{
int n=strlen(a+1);
int m=strlen(b+1);
for(int i=1;i<=m;++i)
{
f[0][i]=f[0][i-1]+(b[i]=='0');
f[1][i]=f[1][i-1]+(b[i]=='1');
}
LL ans=0;
for(int i=1;i<=n;++i)
{
int l=i;
int r=m-(n-i);
if(a[i]=='0')ans+=f[1][r]-f[1][l-1];
else ans+=f[0][r]-f[0][l-1];
}
printf("%lld\n",ans);
}
return 0;
}
/*
【题意】
给你一个模板串a和匹配串b，|a|<=|b|<=2e5，两个串都是01串。
于是，a串之于b串的匹配位置，就有|b|-|a|+1种
我们想问，对于所有的匹配位置，会产生的匹配差异总和是多少。

所以匹配差异，是指——
比如模板串是0011，匹配0101，对于所有'0' match '1' 或者'1' match '0 '，
都产生了1的匹配差异。即这个样例产生了2的匹配差异。

【类型】
水题 前缀和

【分析】
我们直接枚举a串的每个位置，它会对应出b串的一个匹配区间。
我们只要查看区间中有多少字符与其不同即可。
这个可以利用前缀和简单实现。

【时间复杂度&&优化】
O(n+m)

*/