Codeforces Round #336 (Div. 2) 608B Hamming Distance Sum(dp)

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

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.

题目链接:点击打开链接

给出2个字符串a, b. 问字符串b中和字符串a长度相同的子串对应01不同的有多少对.

若a, b两字符串长度相同, 则直接比较. 若字符串长度不同, 记录当前位置以前所有的0, 1个数, 最后累加得到答案.

AC代码:

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"#include "queue"#include "stack"#include "cmath"#include "utility"#include "map"#include "set"#include "vector"#include "list"#include "string"#include "cstdlib"using namespace std;typedef long long ll;const int MOD = 1e9 + 7;const int INF = 0x3f3f3f3f;const int MAXN = 2e5 + 5;char a[MAXN], b[MAXN];ll ans, dp0[MAXN], dp1[MAXN];int main(int argc, char const *argv[]){scanf("%s%s", a + 1, b + 1);int len1 = strlen(a + 1), len2 = strlen(b + 1);if(len1 == len2) {for(int i = 1; i <= len1; ++i)if(a[i] != b[i]) ans++;}else {for(int i = len2; i >= 1; --i)if(b[i] == '0') {dp0[i] = dp0[i + 1] + 1;dp1[i] = dp1[i + 1];}else {dp1[i] = dp1[i + 1] + 1;dp0[i] = dp0[i + 1];}for(int i = 1; i <= len1; ++i)if(a[i] == '0') ans += dp1[i] - dp1[len2 - len1 + i + 1];else ans += dp0[i] - dp0[len2 - len1 + i + 1];}printf("%lld\n", ans);return 0;}