ZOJ3872:Beauty of Array(浙江省赛2015)

最新推荐文章于 2018-11-29 11:56:19 发布

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最新推荐文章于 2018-11-29 11:56:19 发布

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分类专栏：动态规划文章标签： zoj

本文链接：https://blog.youkuaiyun.com/tianwei0822/article/details/76827428

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本文介绍了一种解决特定数组问题的方法：计算所有连续子数组的独特元素之和，并提供了一个AC代码示例。该问题需要使用动态规划技巧来解决。

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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

此题属于序列dp和LIS有异曲同工之妙。（看了题解才恍然大悟，唉）

下面给出AC代码

#include<cstdio>
#include<cstring>
using namespace std;
#define LL long long
int pre[1000005];
LL dp[100005];
int main(){
    int t,n;
    scanf("%d",&t);
    while(t--){
        LL sum = 0,num;
        memset(pre,0,sizeof(pre));
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%lld",&num);
            dp[i]=dp[i-1]+num*(i-pre[num]);///int * int = int 结果可能溢出此题数据比较水 num 不改成long long 也能过
            pre[num]=i;
        }
        for(int i=1;i<=n;i++) sum+=dp[i];
        printf("%lld\n",sum);
    }
}