Codeforces Round #336 (Div. 2) B. Hamming Distance Sum （预处理）

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 s_i is the i-th character of s and t_i 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

0100111

output

3

input

00110110

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.

题意：给你两个串s和t，然后按照题目的方法使s和t中和s一样长的连续子串进行相减，求和

思路：预处理t串前缀和，然后用s串中的每一位数所出现的次数乘以那一位数，然后减去出现的区间中t串的和，取绝对值相加

总结：暴力肯定是不行的，另外，要把ans定义为long long，因为CF不支持，所以用__int64，这里WA了一发

ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 200010#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)using namespace std;char s1[MAXN],s2[MAXN];int sum[MAXN];int main(){int i;while(scanf("%s%s",s1,s2)!=EOF){int len1=strlen(s1);int len2=strlen(s2);sum[0]=s2[0]-'0';for(i=1;i<len2;i++)sum[i]=sum[i-1]+s2[i]-'0';int cnt=len2-len1+1;ll ans=0;//这里wa的一发for(i=0;i<len1;i++){int k=s1[i]-'0';int num=sum[i+len2-len1]-sum[i]+s2[i]-'0';//printf("num=%d\n",num);ans=ans+fabs(k*cnt-num);}printf("%I64d\n",ans);}return 0;}