HDOJ 题目5496 Beauty of Sequence（数学）

Beauty of Sequence

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 363    Accepted Submission(s): 161

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

  

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

 

Sample Output

  

   240
54
144
  

 

Source

BestCoder Round #58 (div.2)

 

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题目大意:所有子序列的和

15013505

2015-10-05 00:15:09

Accepted

5496

1700MS

2816K

782 B

C++

弱渣在努力T^T

ac代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<map>
#define LL __int64
#define mod 1000000007
using namespace std;
LL same[100010];
LL qpow(LL a,int b)
{
	LL ans=1;
	while(b)
	{
		if(b&1)
			ans=(ans*a)%mod;
		a=(a*a)%mod;
		b>>=1;
	}
	return ans;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		int i;
		map<LL,int>mp;
		LL ans=0;
		for(i=1;i<=n;i++)
		{
			LL x;
			scanf("%I64d",&x);
			if(mp.count(x)==0)
			{
				same[i]=0;
				ans=(ans+qpow(2,n-1)*x%mod)%mod;
			}
			else
			{
				ans=(ans+(((qpow(2,i-1)-same[mp[x]])*qpow(2,n-i))%mod*x)%mod)%mod;
				same[i]=same[mp[x]];
			}
			same[i]=(same[i]+qpow(2,i-1))%mod;
			mp[x]=i;
		}
		printf("%I64d\n",(ans+mod)%mod);
	}
}