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

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

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

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

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.

思路：先給出26個字母的權值，再給出一個字符串，然後找出有多少個這樣的子串，滿足頭和尾相同且中間字符的權值和為0（昨天做到這題發現A被個傻逼cha了，無語了，然後就一陣激動沒搞懂這裡啥意思->and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.）

為了算中間為0，只需算出前綴和就好了，用map實現，注意用long long.

AC代碼：

#include 
#include 
#include 
#include 
#include 
#include
#define LL long long 
using namespace std;

int val[30];
char a[100005];
map h[26];

int main() {
	for(int i = 0; i < 26; i++) 
		scanf("%d", &val[i]);
	
	scanf("%s", a);
	int len = strlen(a);
	
	LL ans = 0, sum = 0;
	for(int i = 0; i < len; i++) {
		ans += h[a[i] - 'a'][sum];
		
		sum += val[a[i] - 'a'];
		h[a[i] - 'a'][sum]++;
	}
	
	cout << ans << endl;
	return 0;
}