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

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 n (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

351 2 3 4 541 2 1 353 3 2 1 2

 

Sample Output

24054144

 

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;}