Description
Edward has an array A with N integers. He defines the beauty of an array as the summation of all 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 are multiple test cases. The first line of input contains an integer T indicating the number of 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 in one line.
Sample Input
3 5 1 2 3 4 5 3 2 3 3 4 2 3 3 2
Sample Output
105 21 38
这题直接想到了思路1A了。
假设f(n)表示以a[n]结尾的子区间互异数的和的和。对于f(n-1)，可能会有两部分组成，一部分是包含a[n]这个值的A，一部分是不包含a[n]这个值的B，假设B由p个区间加和。那么可以得到：f(n) = A + B + p*a[n] = f(n-1) + p*a[n]。所以关键是求p。
而由于f(n-1)是以a[n-1]结尾的子区间互异数的和，必然的，假设a[i]是距离a[n]最近的值为a[n]的数，则p = n-i。于是只需要记录最近出现值k的脚标即可。这里采用了map，数据范围不是非常大，也可以直接开数组。
代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <string>
#define LL long long
using 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;
}