Beauty of Array

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

Sample Output

题意

给定一串数字，求所有连续的子序列的和，如果一个数字出现多次，则只计算一次

Solve

去统计每个数字被加的次数即可，注意数据范围

对于一个连续的序列来说，如果有重复的数字，那么后面的数字就可以忽视掉。所以对于一个数字xxx，只需要去计算以这个xxx为结尾和开头的序列，两者相乘即为该数字被加的次数

Code

/*************************************************************************

	 > Author: WZY

	 > School: HPU

	 > Created Time:   2019-04-09 17:07:29

************************************************************************/

#include <bits/stdc++.h>

#define ll long long

#define ull unsigned long long

#define ms(a,b) memset(a,b,sizeof(a))

const int inf=(1<<30);

const ll INF=(1LL*1<<60);

const int maxn=1e6+10;

const int mod=1e9+7;

const int maxm=1e3+10;

using namespace std;

int vis[maxn];

int main(int argc, char const *argv[])

{

	#ifndef ONLINE_JUDGE

	    freopen("in.txt", "r", stdin);

	    freopen("out.txt", "w", stdout);

	#endif

	ios::sync_with_stdio(false);

	cin.tie(0);

	int t;

	cin>>t;

	int n,x;

	ll ans;

	while(t--)

	{

		ms(vis,0);

		ans=0;

		cin>>n;

		for(int i=1;i<=n;i++)

		{

			cin>>x;

			ans+=1LL*x*((i-vis[x])*(n-i+1));

			vis[x]=i;

		}

		cout<<ans<<endl;

	}

	return 0;

}

巴特西

ZOJ 3872： Beauty of Array（思维）