【codechef】Set Difference （灵活题，超时）

Churu is working as a data scientist in Coderpur. He works on a lot of data on the daily basis. One day, he found an interesting problem, which was very easy to solve for small data but was getting more complex with increasing number of data points. So, Churu needs your help in solving this problem.

Given a set S of N non-negative integers (Some integers might occur more than once in the set), find out the value of SETDIFF(S).

Where max(s) represents the maximum value in set s whereas min(s) represents the minimum value in the set s.
As value of SETDIFF(S) can be very large, print it modulo (10⁹ + 7) .

There might be repeated values in the set. For set S = {1,2,2}, consider that first 2 is not same as the second 2 and there will be two different subsets {1,2}. See last sample case for the more clarifications.

Input

• First line of input contains an integer T denoting number of test cases.
• For each test case, first line will contain an integer N denoting number of elements in set S.
• Next line contains N space separated integers denoting the set S.

Output

For each test case, print a single integer representing the answer of that test case.

Note

Two subsets will be called different if there exists an index i such that S[i] occurs in one of the subset and not in another.

Constraints

Subtask #1: 20 points

• 1 ≤ T ≤ 5, 1 ≤ N ≤ 1000, 0 ≤ value in set ≤ 10⁹

Subtask #2: 25 points

• 1 ≤ T ≤ 5, 1 ≤ N ≤ 10⁵, 0 ≤ value in set ≤ 1000

Subtask #3: 55 points

• 1 ≤ T ≤ 5, 1 ≤ N ≤ 10⁵, 0 ≤ value in set ≤ 10⁹

Example

Input:
4
2
1 2
3
1 2 3
4
1 2 3 4
3
1 2 2

Output:
1
6 
23
3

Explanation

For first case answer will be 2-1 = 1.

For the second case:

Subset = {1}, max(s)-min(s) = 0.
Subset = {2}, max(s)-min(s) = 0.
Subset = {3}, max(s)-min(s) = 0.
Subset = {1,2}, max(s)-min(s) = 1.
Subset = {2,3}, max(s)-min(s) = 1.
Subset = {1,3}, max(s)-min(s) = 2.
Subset = {1,2,3}, max(s)-min(s) = 2.
So the output will be 1+1+2+2 = 6.

In the last case, there are three subsets, {1,2}, {1,2} and {1,2,2} having max(s) - min(s) = 1 for each.

http://www.codechef.com/MAY15/problems/SETDIFF

找所有子集合的max-min的和的值

解法：排序，找出所有的i,j(i<j)，sum+=(num[j]-num[j])*pow(j-i-1)。然而，在这里，我们能找出一个巧妙的结论：

式子分解，前半部分系数num[1]=2^0,num[2]=2^0+2^1,num[3]=2^0+2^1+2^2

后半部分系数：num[2]=2^0,num[1]=2^0+2^1,num[0]=2^0+2^1+2^2 =2^3-1。于是可以简化。

#include<iostream>    
#include<algorithm>    
#include<string>    
#include<map>    
#include<vector>    
#include<cmath>    
#include<queue>    
#include<string.h>    
#include<stdlib.h>    
#include<stdio.h>    
#define ll long long 
#define mod 1000000007   
using namespace std;  
ll x[100001];
int y[100001];
int main() {
	y[0]=1;
	for(int i=1;i<100000;++i){
		y[i]=(y[i-1]*2)%mod;
	}
	int t;
	cin>>t;
	while(t--){
		int n;
		cin>>n;
		for(int i=0;i<n;++i)
			cin>>x[i];
		sort(x,x+n);
		ll s=0;
		for(int i=0;i<n;++i){  
			ll p=(x[i]%mod*((y[i]-1)-(y[n-1-i]-1))%mod)%mod;
			s=(s+p)%mod;
		}
		cout<<s<<endl;
	}
	return 0;
}

这是我一开始的代码（20分）：

#include<iostream>  
#include<algorithm>  
#include<string>  
#include<map>  
#include<vector>  
#include<cmath>  
#include<queue>  
#include<string.h>  
#include<stdlib.h>  
#include<stdio.h>
#define ll long long  
#define mod 1000000007
using namespace std;
ll x[100001];
ll y[100001];
int main(){
	y[0]=1;
	for(int i=1;i<100000;++i)
		y[i]=(y[i-1]*2)%mod;
	int t;
	scanf("%d",&t);
	while(t--){
		int n;
		scanf("%d",&n);
		for(int i=0;i<n;++i)
			scanf("%I64d",&x[i]);
		sort(x,x+n);
		ll s=0;
		for(int i=0;i<n;++i){
			for(int j=i+1;j<n;++j){
				ll p=(x[j]-x[i])%mod;
				p=(p*y[j-i-1])%mod;
				s=(s+p)%mod;
			}
		}
		printf("%I64d\n",s);
	}
	return 0;
}