Andyqsmart ACM Learning process--zoj3872 Beauty of Array (recursive)

Description

Edward has a array A with N integers. He defines the beauty of an array as the summation of a distinct integers in the array. Now Edward wants to know the summation of the beauty of all contiguous subarray of the array A .

Input

There is multiple test cases. The first line of input contains an integer indicating the number of the T test cases. For each test case:

The first line contains an integer N (1 <= N <= 100000), which indicates the size of the array. The next line contains N positive integers separated by spaces. Every integer is no larger than 1000000.

Output

For each case, print the answer on one line.

Sample Input

351 2 3 4 532 3 342 3 3 2

Sample Output

1052138

This problem directly thought of the idea of 1 a.

Suppose that f (n) represents the and of the number of sub-intervals that end with a[n]. For f (n-1), there may be two parts, part A of the value of A[n], part B that does not contain the value of a[n], assuming B is added by the P interval. Then you can get: F (N) = A + B + p*a[n] = f (n-1) + p*a[n]. So the key is to ask for P.

And since F (n-1) is the number of sub-interval reciprocal numbers ending with a[n-1], it is necessary to assume that A[i] is the nearest value of A[n] for the distance a[n], then P = n-i. So you only need to record the most recent value K's foot mark. Map is used here, the data range is not very large, you can open the array directly.

Code:

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath > #include <algorithm> #include <set> #include <map> #include <queue> #include <string> #define LL Long longusing namespace Std;int main () {    //freopen ("test.in", "R", stdin);    int T, n, V, K;    LL ans, F;    scanf ("%d", &t);    for (int times = 0; times < T; ++times)    {        scanf ("%d", &n);        Map<int, int> s;        Ans = 0;        f = 0;        for (int i = 1; I <= n; ++i)        {            scanf ("%d", &v);            f = f + (I-s[v]) *v;            S[V] = i;            Ans + = f;        }        printf ("%lld\n", ans);    }    return 0;}

Andyqsmart ACM Learning process--zoj3872 Beauty of Array (recursive)