【codeforces】Round #294 (Div. 2) D. A and B and Interesting Substrings

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本文介绍了一种算法，用于计算字符串中满足特定条件的子串数量，这些子串需首尾字符相同且除去首尾的字符权值之和为零。通过前缀和与映射表的方法，实现高效计算。

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A and B are preparing themselves for programming contests.

After several years of doing sports programming and solving many problems that require calculating all sorts of abstract objects, A and B also developed rather peculiar tastes.

A likes lowercase letters of the Latin alphabet. He has assigned to each letter a number that shows how much he likes that letter (he has assigned negative numbers to the letters he dislikes).

B likes substrings. He especially likes the ones that start and end with the same letter (their length must exceed one).

Also, A and B have a string s. Now they are trying to find out how many substrings t of a string s are interesting to B (that is, t starts and ends with the same letter and its length is larger than one), and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.

Naturally, A and B have quickly found the number of substrings t that are interesting to them. Can you do it?

Input

The first line contains 26 integers xa, xb, ..., xz ( - 105 ≤ xi ≤ 105) — the value assigned to letters a, b, c, ..., z respectively.

The second line contains string s of length between 1 and 105 characters, consisting of Lating lowercase letters— the string for which you need to calculate the answer.

Output

Print the answer to the problem.

Sample test(s)

input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
xabcab

output
2

input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
aaa

output
2

Note

In the first sample test strings satisfying the condition above are abca and bcab.

In the second sample test strings satisfying the condition above are two occurences of aa.

题意：

为每一个字母分配一个权值，求满足以下条件的子串个数： 1子串首尾字母相同，长度大于一 2.子串除首尾外的字母权值之和为0

思路：

1.与连续和相关的问题一般会涉及到前缀和（sum[i]），通过前缀和O（n）的预处理可以将区间和的判断降到O（1）时间复杂度

2.这里求相同字母之间的连续和为0的子串数，即求相同字母具有相同的前缀和的对数，比如序列中存在两个a，分别位于str[3]和str[7]，如果前缀和sum[3]==sum[7]，即意味着str的子串3~7中的连续和为0，满足题目要求，因此可以采用dp方式，使用数组dp[x][y]表示字母x前缀和为y的个数

3.由于y可能会很大，可能导致数组浪费或开不下，所以这里直接使用map数组

4.在dp过程维护map数组的时候顺带维护一个前缀和cnt即可

代码如下：

/*
	Author : _L
*/
#include<cstdio>
#include<cmath>
#include<iomanip>
#include<cstring>
#include<cstdlib>
#include<ctime>
#include<iostream>
#include<sstream>
#include<fstream>
#include<algorithm>
#include<vector>
#include<list>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<string>
#include<bitset>
#include<functional>
#include<utility>

using namespace std;

typedef long long ll;
const int maxn=1e5+50;
int v[30];
map<ll,ll>MAP[30];
char str[maxn];
int m,n;


int main()
{
    int i,j,k,tt,a,b,len;
    ll ans,cnt;

    for(i=0;i<26;i++)
    {
        scanf("%d",&v[i]);
    }
    scanf("%s",str);

    len=strlen(str);
    ans=cnt=0;
    for(i=0;i<len;i++)
    {
        ans+=MAP[str[i]-'a'][cnt];
        cnt+=v[str[i]-'a'];
        MAP[str[i]-'a'][cnt]++;
    }
    printf("%I64d\n",ans);


    return 0;
}