zoj3872 Beauty of Array（dp）

最新推荐文章于 2018-11-29 11:56:19 发布

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本博客探讨了一个数组中连续子序列的美丽度之和的计算方法，美丽度定义为子序列中唯一整数的总和。通过实例演示和代码实现，详细解释了如何遍历数组并计算所有子序列的美丽度。

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

105
21
38
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5520
用map超时，简化后仍超时，当时一直往这个方向钻研了几个小时，队友说可能是dp被我一口否决。。结果可想而知╮(╯▽╰)╭
dp[i]表示长度为i的数组所有子序列总和。
#include<iostream>  
#include<algorithm>  
#include<string>  
#include<map>  
#include<string.h>  
#include<vector>  
#include<cmath>  
#include<stdlib.h>  
#include<cstdio>  
#define ll long long  
using namespace std;
int x[1000001]; 
ll dp[100001];
int main(){
	int t;
	cin>>t;
	while(t--){
		memset(x,0,sizeof(x));
		int n,a;
		cin>>n;
		dp[0]=0;
		for(int i=1;i<=n;++i){
			cin>>a;
			dp[i]=dp[i-1]+(i-x[a])*a;
			x[a]=i;
		}
		ll s=0;
		for(int i=1;i<=n;++i)
			s+=dp[i];
		cout<<s<<endl;
	}
	return 0;
}