2017ecjtu-summer training #2 CodeForces 608B

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 |.

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.

Print a single integer — the sum of Hamming distances between a and all contiguous substrings of b of length | a |.

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: 定义两个字符串之间的距离是所有对应的每个字符相减绝对值的和, 给定a和b两个字符串, 求b中与a长度相同的所有子串与a字符串距离之和.

分析: 多写一些样例,就可以归纳出来, 其实就是a字符串中的第一个元素,分别 与b字符串中的第一个元素至第|b|-|a|+1个元素求距离的和, 再加上a字符串中的第2个元素,分别 与b字符串中的第2个元素至第|b|-|a|+2个元素求距离的和......直到a字符串中的第|a|个元素与b字符串中的第|a|个元素至第|b|个元素求距离的总和. 由于求距离很像求异或和, 因此可以先用两个数组分别保存b字符串的0个数的前缀和和1个数的前缀和, 遍历a字符串的时候 ,如果字符是1 ,那么求相应区间0的的个数,  如果字符是0 ,那么求相应区间1的的个数.

AC代码

 #include<cstdio>

 #include<iostream>

 #include<cstring>

 using namespace std;

 #define maxn 222222

 typedef long long ll;

 char a[maxn],b[maxn];

 int res1[maxn],res0[maxn];

 int main()

 {

     while(~scanf("%s%s",a+,b+))

     {

         int la=strlen(a+),lb=strlen(b+);

         res1[]=res0[]=;

         for(int i=; i<=lb; i++)

         {

             if(b[i]=='')

             {

                 res1[i]=res1[i-]+;

                 res0[i]=res0[i-];

             }

             else

             {

                 res1[i]=res1[i-];

                 res0[i]=res0[i-]+;

             }

         }

         ll ans=;

         for(int i=; i<=la; i++)

             if(a[i]=='')

                 ans+=res0[lb-(la-i)]-res0[i-];

             else

                 ans+=res1[lb-(la-i)]-res1[i-];

         printf("%I64d\n",ans);

     }

     return ;

 }