ZOJ3872 Beauty of Array【DP】

Beauty of Array

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

3
5
1 2 3 4 5
3
2 3 3
4
2 3 3 2

Sample Output

105
21
38

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

问题链接：ZOJ3872 Beauty of Array。

问题简述：参见上文。

题意有点难懂。

给定n个数的序列，求n个数的各个子序列不重复元素和的和。

例如3个元素的序列2,3,3，各个子序列就是{2}，{2，3}，{2，3，3}，{3}，{3，3}，{3}。各个子序列中去掉重复元素后求和，总和就是2+(2+3)+(2+3)+3+3+3=21。

问题分析：这个题需要用动态规划来解。递推函数dp[i]=(i-pos[ai])*ai，其中pos[ai]=i即上一次ai出现的位置。

程序说明：（略）

题记：（略）

AC的C++语言程序如下：

/* ZOJ3872 Beauty of Array */

#include <iostream>
#include <stdio.h>
#include <string.h>

const int N = 1000000;
long long pos[N+1];

using namespace std;

int main()
{
    int t, n, a;
    scanf("%d", &t);
    while(t--) {
        memset(pos, 0, sizeof(pos));

        scanf("%d", &n);
        long long sum = 0, dp = 0;
        for(int i=1; i<=n; i++) {
            scanf("%d", &a);
            dp += (i - pos[a]) * a;
            sum += dp;
            pos[a] = i;
        }

        printf("%lld
", sum);
    }

    return 0;
}