D - Beauty of Array ZOJ - 3872

最新推荐文章于 2021-02-22 21:40:33 发布

原创最新推荐文章于 2021-02-22 21:40:33 发布 · 337 阅读

0 ·

CC 4.0 BY-SA版权

策略专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一种通过计算序列中所有连续子数组的唯一整数之和来解决特定问题的方法。利用动态规划技术和额外的数据结构简化了计算过程，提高了算法效率。

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

题意：给定一个数字序列，把这个序列分成连续的串，在这个序列中如果一个数之出现一次，则把这些数加起来，如果多次出现

则之记录一次，求出所有的序列计数之和；

解题思路：先用s存下当前的数这个题可以简单用下dp和一个1e6的数组k，dp记录到当前为之计数之和的个数，数组k记录着

s[i]的数上次出现的位置则递推方程为dp[i]=dp[i-1]+s[i]*(i-k[s[i]]);

#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
#include <cmath>
#include <algorithm>
#define pb push_back
#define mk make_pair
typedef long long ll;
using namespace std;
const ll inf = 1e5+100;
const ll INF = 1e6+1;
ll s[inf],dp[inf],k[INF];
int main()
{
    ll t;scanf("%lld",&t);
    while(t--)
    {
        ll n;scanf("%lld",&n);
        memset(k,0,sizeof(k));
        for(int i=1;i<=n;i++)scanf("%lld",&s[i]);
        s[0]=0;dp[0]=0;
        for(int i=1;i<=n;i++)
        {
            dp[i]=s[i]*(i-k[s[i]])+dp[i-1];
            k[s[i]]=i;
        }
        /*for(int i=1;i<=n;i++)
            cout<<dp[i]<<" ";
        cout<<endl;*/
        ll ans=0;
        for(int i=1;i<=n;i++)
            ans+=dp[i];
        printf("%lld\n",ans);
    }

    return 0;
}