codeforces #336 div 2 B. Hamming Distance Sum

http://codeforces.com/contest/608/problem/B 题意:给定两个字符串a,b,问b中的每个连续的长度为a的子串与a的哈密顿距离的和是多少。哈密顿距离是对应位置的字符的差的绝对值的和。由于是01串,也就是字符不同的位置数。 思路:类似前缀和。0和1分别搞。注意开long long

 1/* ***********************************************
 2Author :111qqz
 3Created Time :2015年12月24日 星期四 00时32分33秒
 4File Name :code/cf/#336/B.cpp
 5************************************************ */
 6
 7#include <cstdio>
 8#include <cstring>
 9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=3E5+7;
34int la,lb;
35char a[N],b[N];
36int sum0[N],sum1[N];
37int main()
38{
39	#ifndef  ONLINE_JUDGE 
40	freopen("code/in.txt","r",stdin);
41  #endif
42	ms(sum0,0);
43	ms(sum1,0);
44	scanf("%s",a);
45	scanf("%s",b);
46	la = strlen(a);
47	lb = strlen(b);
48	int sum = 0 ;
49	if (la==1)
50	{
51	    for ( int i = 0 ;  i < lb ; i++)
52	    {
53		if (b[i]!=a[0]) sum++;
54	    }
55	    cout<<sum<<endl;
56	    return 0;
57
58	}
59
60	for ( int i = 0 ; i < lb ; i++)
61	{
62	    if (b[i]=='0')
63	    {
64		sum0[i+1] = sum0[i]+1;
65		sum1[i+1] = sum1[i];
66	    }
67	    else
68	    {
69		sum1[i+1] = sum1[i] +1;
70		sum0[i+1] = sum0[i];
71	    }
72	}
73//	for ( int i = 1 ; i <= lb ; i++)
74//	    cout<<sum0[i]<<" "<<sum1[i]<<endl;
75	LL ans =  0;
76	for (int i= 0 ;  i < la ; i++)
77	{
78	    if (a[i]=='0')
79	    {
80		ans+=LL(sum1[lb-la+i+1]-sum1[i]);
81	    }
82	    else
83	    {
84		ans+=LL(sum0[lb-la+i+1]-sum0[i]);
85	    }
86//	    cout<<ans<<endl;
87	}
88
89	cout<<ans<<endl;
90
91  #ifndef ONLINE_JUDGE  
92  fclose(stdin);
93  #endif
94    return 0;
95}