Beauty of Array(思维)

Beauty of Array

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 Npositive 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
题解：找集合A中连续的连续子集内的不同元素的和；思维转化成每个元素在当前连续子集中需要加的个数就好了；
代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<vector>
using namespace std;
const int MAXN = ;
int num[MAXN];
int pre[MAXN];
typedef long long LL;
LL sum[MAXN];
int main()
{
    int T, N;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &N);
        memset(pre, , sizeof(pre));
        memset(sum, , sizeof(sum));
        int x;
        LL ans = ;
        for(int i = ; i <= N; i++){
            scanf("%d", &x);
            sum[i] = sum[i - ] + (i - pre[x]) * x;//i - pre[i]代表x用到了多少次；
            pre[x] = i;
            ans += sum[i];
        }
        printf("%lld\n", ans);
    }
    return ;
}