zoj3872——Beauty of Array（数学）

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

351 2 3 4 532 3 342 3 3 2

Sample Output

1052138

dp[i]=dp[i-1]+a*(i-pre[a]);

求相邻数组的总和，其中数组中不能有重复的数字。设dp[i]为长度为i的数组的总和，用pre[a]记录数字a最后出现的位置，以1，2,3,2为例

i=1,a=1

dp[i]=1;

i=2,a=2

dp[i]=2+(1+2);

i=3,a=3

dp[i]=3+(2+3)+(1+2+3);

i=4,a=2,pre[2]=2

dp[i]=2+(3+2)+(2+3+{2})+(1+2+3+{2});

#include<iostream>#include<cstdio>#include<cstring>#define MAXN 1000010using namespace std;long long pre[MAXN],dp[MAXN],sum;int main(){    int t,n,i,a;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        memset(pre,0,sizeof(pre));        dp[0]=0;        for(i=1;i<=n;++i)        {            cin>>a;            dp[i]=dp[i-1]+a*(i-pre[a]);            pre[a]=i;        }        sum=0;        for(i=1;i<=n;++i)            sum+=dp[i];        cout<<sum<<endl;    }    return 0;}