hdoj 5496 Beauty of Sequence

Beauty of Sequence

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 767    Accepted Submission(s): 348

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

 

给出由n个数组成的序列，求出所有去重后的子序列的和；

子序列的末尾元素有n种，a[1],a[2],a[3]...a[n]；

设f[i]为前i个元素组成序列的所有子序列的和（去重后）；

设add表示以元素a[i]结尾的子序列的和（去重后）；

则f[i]=f[i-1]+add；

设sum表示以元素a[i]结尾的子序列的和（没有去重）；

则sum=F[i-1](不管怎样F[i-1]一定会有)+2^(i-1)*a[i](我们贪心的认为前面所有方案都成立，然后去重)。

设重复包含a[i]的序列有same[a[i]]个，那么add=sum-same[a[i]]*a[i]。

可以选择用map建立映射<LL, LL>same计算same[[ai]]；

same[a[i]]=same[a[i]]+2^(i-1)；

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#include<map>
#define LL long long
#define MAXN 100000+10
#define MOD 1000000007
LL a[MAXN];
LL f[MAXN];//f[i]为前i个元素组成序列的所有子序列的和（去重后） 
map<LL,LL>same;
LL pow_mod(LL a,LL b)//a的b次方 
{
	LL ans=1;
	while(b)
	{
		if(b&1)
		ans=ans*a%MOD;
		a=a*a%MOD;
		b=b/2;
	}
	return ans;
}
int main()
{
	int T,n,i;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		same.clear();
		f[0]=0;
		for(i=1;i<=n;i++)
		{
			scanf("%lld",&a[i]);
			LL sum=(f[i-1]+pow_mod(2,i-1)*a[i]%MOD)%MOD;
			//sum表示以元素a[i]结尾的子序列的和（没有去重） 
			LL add=(sum+MOD-same[a[i]]*a[i]%MOD)%MOD;
			//add表示以元素a[i]结尾的子序列的和（去重后） 
			f[i]=(f[i-1]+add)%MOD;
			same[a[i]]=(same[a[i]]+pow_mod(2,i-1))%MOD;
			//重复包含a[i]的序列有same[a[i]]个 
		}
		printf("%lld\n",f[n]);
	}
	return 0;
}