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

原题链接：

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.

题意：求满足两个条件的长度大于一的子串个数。条件一：首位字母相同，条件二：除了首位字母其它字母权值为0。

方法：

①求出前缀和sum[i]；

②从开头枚举，在位置i，以它结尾的可行子串个数为map[它][sum[i-1]]，再把sum[i]存入相应的map。

代码：

#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;
}