Codeforces 809A - Do you want a date?(数学+排序)
Leha decided to move to a quiet town Vi?kopolis, 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 Vi?kopolis.
Let's denote the coordinate system on this street. Besides let'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'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?+?7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
InputThe first line contains one integer?n?(1?≤?n?≤?3·105)?denoting the number of hacked computers.
The second line contains?n?integers?x1,?x2,?...,?xn?(1?≤?xi?≤?109)?denoting the coordinates of hacked computers. It is guaranteed that allxi?are distinct.
OutputPrint a single integer?— the required sum modulo?109?+?7.
Examples input 24 7 output 3 input 3
4 3 1 output 9 Note
There are three non-empty subsets in the first sample test:,??and?. The first and the second subset increase the sum by?0and the third subset increases the sum by?7?-?4?=?3. In total the answer is?0?+?0?+?3?=?3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer:?,?,?,?. In total the sum is?(4?-?3)?+?(4?-?1)?+?(3?-?1)?+?(4?-?1)?=?9.
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題意:給一個長度為n的數組(數組元素沒有重復,可以看成一個集合),求這個集合的子集中最大值和最小值的差,然后求和。
我們知道n個元素的集合的子集為2^n個,把數組從小到大排序后,任意的第i個元素和第j個元素(i<j)可以算出關系為2^(j-i-1)*(a[j]-a[i]),然后根據這個式子可以寫出代碼。
代碼:
for(i=1;i<=n;i++){for(j=i+1;j<=n;j++){ans+=b[j-i-1]*(a[j]-a[i])%mod;ans%=mod;}}//數組b為2^(n-1)?
然而這個算法的時間復雜度為O(n^2),很明顯地會超時,不符合題目要求。
然后對a[i]這個元素進行觀察和推算,可以推出差值為(2^(i-1)-2^(n-i))*a[i]。
AC代碼:
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#include<iostream> #include<algorithm> using namespace std; const int maxn = 3e5 + 5; typedef long long LL; LL a[maxn], b[maxn]; #define mod 1000000007 int main() {int n;while (cin >> n){int i;b[0] = 1;for (i = 1; i <= n; i++){cin >> a[i];b[i] = (2 * b[i - 1]) % mod;}sort(a + 1, a + 1 + n);LL ans = 0;for (i = 1; i <= n; i++)ans = (ans + (b[i - 1] - b[n - i])*a[i]) % mod;cout << ans << endl;}return 0; }?
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轉載于:https://www.cnblogs.com/orion7/p/6917395.html
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