Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let&＃39;s denote the coordinate system on this street. Besides let&＃39;s number all the hacked computers with integers from 1 to n. So the i-th hacked computer is located at the point xi. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let&＃39;s denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression 
. Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, 
. Since the required sum can be quite large Noora asks to find it modulo 109 &＃43; 7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Output
Print a single integer — the required sum modulo 109 &＃43; 7.
题意&＃xff1a;给一个数列&＃xff0c;求其所有子集内最大值减最小值的和。 思路&＃xff1a;先排序&＃xff0c;固定左端点i&＃xff0c;枚举右端点j&＃xff0c;那么为答案贡献a[j]-a[i]的有2^(j-i-1)个子集&＃xff0c;但是On^2必然超时&＃xff0c;考虑到a[j]-a[i]&＃xff0c;a[j&＃43;1]-a[i&＃43;1]这些有共同之处&＃xff0c;列式子化简可发现和前缀和和后缀和有关。
# include
using namespace std;typedef long long LL;
const int maxn &＃61; 3e5&＃43;3;
const LL mod &＃61; 1e9&＃43;7;
LL a[maxn]&＃61;{0}, b[maxn]&＃61;{0}, c[maxn];LL qmod(LL a, LL b)
{LL pow &＃61; a, ans &＃61; 1;while(b){if(b&1) ans &＃61; ans*pow%mod;pow &＃61; pow*pow%mod;b >>&＃61; 1;}return ans;
}int main()
{int n;scanf("%I64d",&n);for(int i&＃61;1; i<&＃61;n; &＃43;&＃43;i) scanf("%I64d",&c[i]);sort(c&＃43;1, c&＃43;1&＃43;n);for(int i&＃61;1; i<&＃61;n; &＃43;&＃43;i) a[i] &＃61; (a[i-1]&＃43;c[i])%mod;for(int i&＃61;n; i>&＃61;1; --i) b[i] &＃61; (b[i&＃43;1]&＃43;c[i])%mod;LL ans &＃61; 0;for(int i&＃61;1; i<&＃61;n; &＃43;&＃43;i)ans &＃61; (ans &＃43; ((b[n-i&＃43;1]-a[i]&＃43;mod)%mod)*qmod(2LL, (LL)i-1))%mod;printf("%I64d\n",ans);return 0;
}
, 
and 
. The first and the second subset increase the sum by 
, 
, 
, 
. In total the sum is 
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