hdoj 5496 Beauty of Sequence 【求序列所有子序列（去重后）的和】【好题】

本文探讨了一道算法题目，要求计算给定整数序列的所有子序列去重后的和，并提供了解题思路与AC代码实现。

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Beauty of Sequence

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 304 Accepted Submission(s): 135

Problem Description

Sequence is beautiful and the beauty of an integer sequence is defined as follows: removes all but the first element from every consecutive group of equivalent elements of the sequence (i.e. unique function in C++ STL) and the summation of rest integers is the beauty of the sequence.

Now you are given a sequence

A of

n integers

{a1,a2,...,an}. You need find the summation of the beauty of all the sub-sequence of

A. As the answer may be very large, print it modulo

109+7.

Note: In mathematics, a sub-sequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example

{1,3,2} is a sub-sequence of

{1,4,3,5,2,1}.

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

(1≤n≤105), indicating the size of the sequence. The following line contains

n integers

a1,a2,...,an, denoting the sequence

(1≤ai≤109).

The sum of values

n for all the test cases does not exceed

2000000.

Output

For each test case, print the answer modulo

109+7 in a single line.

Sample Input

Sample Output

240
54
144

序列去重后的和 —— 如 1 2 2 2 3 ，去重后和为1 + 2 + 3 = 6。

题意：给你一个N个数组成的序列，求出所有子序列（去重后）的和。

思路：考虑每个元素作为第一个元素的贡献，然后计数即可。

AC代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cstdlib>
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
const int MAXN = 1e5 + 10;
typedef pair<int, int> pii;
const int INF = 1e9 + 10;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
using namespace std;
int a[MAXN];
LL f[MAXN];;
map<int, LL> res;
int main()
{
    f[0] = 1;
    for(int i = 1; i <= MAXN - 1; i++) {
        f[i] = f[i - 1] * 2 % MOD;
    }
    int t; scanf("%d", &t);
    while(t--) {
        int n; scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        LL ans = 0; LL sum = 0; res.clear();
        for(int i = 1; i <= n; i++) {
            add(sum, f[i - 1]); add(res[a[i]], f[i - 1]);
            add(ans, f[n - i] * a[i] % MOD * ((sum - res[a[i]] + MOD) % MOD + 1) % MOD);
        }
        printf("%lld\n", ans);
    }
    return 0;
}