Codeforces Round #294 (Div. 2) D. A and B and Interesting Substrings

原題鏈接：

http://codeforces.com/contest/519/problem/D

D. A and B and Interesting Substrings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

題意：求滿足兩個條件的長度大於一的子串個數。條件一：首位字母相同，條件二：除了首位字母其它字母權值爲0。

方法：

①求出前綴和sum[i]；

②從開頭枚舉，在位置i，以它結尾的可行子串個數爲map[它][sum[i-1]]，再把sum[i]存入相應的map。

代碼：

#include "stdio.h"
#include "iostream"
#include "string.h"
#include "stdlib.h"
#include "algorithm"
#include "math.h"
#include "map"
#include "string"
using namespace std;

int num[27],len;
char in[100100];
long long ans=0,sum[100100]={0};
map <long long,int> p[27];

int main()
{
	for(int i=0;i<26;i++)
	{
		scanf("%d",&num[i]);
		p[i].clear();
	}
	scanf("%s",in);
	len=strlen(in);
	sum[0]=num[in[0]-'a'];
	for(int i=1;i<len;i++)
	{
		sum[i]+=sum[i-1]+num[in[i]-'a'];
	//	printf("%d %d\n",i,sum[i]);
	}
	p[in[0]-'a'][sum[0]]=1;
	for(int i=1;i<len;i++)
	{
		ans+=p[in[i]-'a'][sum[i-1]];
	//	printf("%c %lld %d %lld\n",in[i],sum[i-1],i,ans);
		p[in[i]-'a'][sum[i]]++;
	}
	printf("%I64d\n",ans);
	return 0;
}