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?
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.
Print the answer to the problem.
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
2
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
2
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：在给的一个串，每个字母有固定的权值求满足如下条件的子串的数目
1 该子串的首字母和尾字母相同
2 该串除了首尾字母外，其他字母的权值和为0
思路：维护一个前缀和，用了26个hash分别记录了26种字母每个的前缀和，如果遇到了一个'a'，就检查之前有几个和他的前缀和一样的'a'，在和上加上去，并维护前缀和。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<string> 6 #include<queue> 7 #include<algorithm> 8 #include<map> 9 #include<iomanip> 10 #include<climits> 11 #include<string.h> 12 #include<cmath> 13 #include<stdlib.h> 14 #include<vector> 15 #include<stack> 16 #include<set> 17 using namespace std; 18 #define INF 1000000007 19 #define MAXN 40010 20 #define Mod 1000007 21 #define N 10007 22 #define NN 30 23 #define sigma_size 3 24 const int maxn = 6e5 + 10; 25 using namespace std; 26 typedef long long LL; 27 28 int h[30]; 29 map<LL, LL> m[30]; 30 string s; 31 LL cnt, k; 32 33 int main() 34 { 35 for (int i = 0; i < 26; ++i) 36 cin >> h[i]; 37 cin >> s; 38 for (int i = 0; i < s.length(); ++i) { 39 cnt += m[s[i] - 'a'][k]; 40 k += h[s[i] - 'a']; 41 m[s[i] - 'a'][k]++; 42 } 43 cout << cnt << endl; 44 //system("pause"); 45 return 0; 46 }
Anya has bought a new smartphone that uses Berdroid operating system. The smartphone menu has exactly n applications, each application has its own icon. The icons are located on different screens, one screen contains k icons. The icons from the first to the k-th one are located on the first screen, from the (k + 1)-th to the 2k-th ones are on the second screen and so on (the last screen may be partially empty).
Initially the smartphone menu is showing the screen number 1. To launch the application with the icon located on the screen t, Anya needs to make the following gestures: first she scrolls to the required screen number t, by making t - 1 gestures (if the icon is on the screen t), and then make another gesture — press the icon of the required application exactly once to launch it.
After the application is launched, the menu returns to the first screen. That is, to launch the next application you need to scroll through the menu again starting from the screen number 1.
All applications are numbered from 1 to n. We know a certain order in which the icons of the applications are located in the menu at the beginning, but it changes as long as you use the operating system. Berdroid is intelligent system, so it changes the order of the icons by moving the more frequently used icons to the beginning of the list. Formally, right after an application is launched, Berdroid swaps the application icon and the icon of a preceding application (that is, the icon of an application on the position that is smaller by one in the order of menu). The preceding icon may possibly be located on the adjacent screen. The only exception is when the icon of the launched application already occupies the first place, in this case the icon arrangement doesn't change.
Anya has planned the order in which she will launch applications. How many gestures should Anya make to launch the applications in the planned order?
Note that one application may be launched multiple times.
The first line of the input contains three numbers n, m, k (1 ≤ n, m, k ≤ 105) — the number of applications that Anya has on her smartphone, the number of applications that will be launched and the number of icons that are located on the same screen.
The next line contains n integers, permutation a1, a2, ..., an — the initial order of icons from left to right in the menu (from the first to the last one), ai — is the id of the application, whose icon goes i-th in the menu. Each integer from 1 to n occurs exactly once among ai.
The third line contains m integers b1, b2, ..., bm(1 ≤ bi ≤ n) — the ids of the launched applications in the planned order. One application may be launched multiple times.
Print a single number — the number of gestures that Anya needs to make to launch all the applications in the desired order.
8 3 3
1 2 3 4 5 6 7 8
7 8 1
7
5 4 2
3 1 5 2 4
4 4 4 4
8
In the first test the initial configuration looks like (123)(456)(78), that is, the first screen contains icons of applications 1, 2, 3, the second screen contains icons 4, 5, 6, the third screen contains icons 7, 8.
After application 7 is launched, we get the new arrangement of the icons — (123)(457)(68). To launch it Anya makes 3 gestures.
After application 8 is launched, we get configuration (123)(457)(86). To launch it Anya makes 3 gestures.
After application 1 is launched, the arrangement of icons in the menu doesn't change. To launch it Anya makes 1 gesture.
In total, Anya makes 7 gestures.
题意：n个数分布在几个格子中，一个格子最多装k个数，可以不装，每次到达一个数，把这个数跟前边的数的位置换一下（除非前边没有别的数），求次数
思路：对数跟下边的玩味 hash存下表，每次到达一个数，跟前边的数交换一次下标
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<string> 6 #include<queue> 7 #include<algorithm> 8 #include<map> 9 #include<iomanip> 10 #include<climits> 11 #include<string.h> 12 #include<cmath> 13 #include<stdlib.h> 14 #include<vector> 15 #include<stack> 16 using namespace std; 17 #define INF 1000000007 18 #define MAXN 40010 19 #define Mod 1000007 20 #define N 10007 21 #define NN 30 22 #define sigma_size 3 23 const int maxn = 6e5 + 10; 24 using namespace std; 25 typedef long long LL; 26 27 int n, m, k; 28 int x; 29 int a[100100]; 30 int b[100100]; 31 32 int main() 33 { 34 cin >> n >> m >> k; 35 for (int i = 1; i <= n; ++i) { 36 cin >> a[i]; 37 b[a[i]] = i; 38 } 39 LL re = 0; 40 for (int i = 0; i < m; ++i) { 41 cin >> x; 42 re += (b[x]-1) / k + 1; 43 //cout << re << endl; 44 x = b[x]; 45 if (x>1) 46 { 47 swap(a[x - 1], a[x]); 48 b[a[x - 1]] = x - 1; 49 b[a[x]] = x; 50 } 51 } 52 cout << re << endl; 53 //system("pause"); 54 return 0; 55 }