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 integerT, 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

 

Source

BestCoder Round #58 (div.2)

 

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题目大意:所有子序列的和

150135052015-10-05 00:15:09Accepted54961700MS2816K782 BC++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 1000000007using 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);}}