ZOJ 3872 Beauty of Array

最新推荐文章于 2020-06-22 12:17:52 发布

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#ZOJ #3872 #Beauty of Array

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探讨如何通过动态规划高效解决数组中所有连续子数组的唯一元素求和问题，给出清晰的实现思路与代码示例。

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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

105
21
38

Author: LIN, Xi
Source: The 12th Zhejiang Provincial Collegiate Programming Contest

题目票两张数组，真漂亮啊，三个小时，宝宝硬是没做出来，尴尬，后来看他们题解，恩、就是那么回事，但是宝宝就是做不出来，这咋办也知道暴力的逆过程就是动态，但是宝宝不知道公式，这咋办，问队长，绝对是多做题就好哦啊。

题意：给你一个数组，让你找出来他所有连续子串，对每个字符串，求其所有元素加起来的和（对于重复的只保留一个）

题解：

说不清楚

//输入 1 2 3

//dp 1 5 14

//sum 1 6 20

//a[i] 1 2 3

接下来就是看代码，记得用long long

#include <iostream>
#include <cstring>
#include<stdio.h>
using namespace std;

const int MAXN = 1e5 + 10;

int main()
{
    int a[MAXN];
    int t, n;
    scanf ("%d", &t);
    while (t--)
    {
        memset (a, 0, sizeof (a));
        scanf ("%d", &n);
        int x;
        long long dp = 0, sum = 0;

        for (int i=1; i<=n; ++i)
        {
            scanf ("%d", &x);
            dp = (i - a[x]) * x + dp;
            sum += dp;
            a[x] = i;
        }

        printf ("%lld\n", sum);

    }
    return 0;
}