Codeforces Round #336 (Div. 2) B 暴力

連結：戳這裡

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|.

Examples
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 is 1 + 0 + 1 + 1 = 3.

The second sample case is described in the statement.

題意：

給出兩串01串s,p 要求p中所有長度為len(s)的串與s串相減絕對值，然後累加起來並輸出

思路：

嗯 維護一下字首和然後算貢獻

程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
string s,p;
int sum[200100];
int a[200100];
int b[200100];
int main(){
    cin>>s>>p;
    int n=s.size(),m=p.size();
    for(int i=1;i<=n;i++){
        b[i]=s[i-1]-'0';
        b[i]*=(m-n+1);
    }
    sum[0]=0;
    for(int i=1;i<=m;i++){
        int a=p[i-1]-'0';
        sum[i]=sum[i-1]+a;
    }
    int cnt=0;
    for(int i=m-n+1;i<=m;i++){
        a[++cnt]=sum[i]-sum[i-(m-n+1)];
    }
    ll ans=0;
    for(int i=1;i<=n;i++){
        ans+=abs(b[i]-a[i]);
    }
    cout<<ans<<endl;
    return 0;
}
/*
1001101001101110101101000
01111000010011111111110010001101000100011110101111
*/