【Codeforces 809 A. Do you want a date?】

本文介绍了一个算法问题,即计算所有非空子集中计算机间最大距离之和,并给出了解决方案。利用组合数学的方法,通过计算每段连续区间被使用的次数来得出最终结果。

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A. Do you want a date?
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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’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.
Input
The 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 all xi are distinct.
Output
Print a single integer — the required sum modulo 109 + 7.
Examples
Input
2
4 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 0 and 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.

比较不好想的题,求ai 和 aj 的距离实际上会把ai 和 aj 中间点的距离都会算上,实际上就是求任意两个相邻的点这段距离用到了多少次,这段距离用的次数不难想:cut =( C(1,i) + C(2,i)+…C(i,i) ) * (C(1,n - i) + C(2,n - i) + …..+ C(n - i,n - i)),而 C(0,n) + C(1,n) + ….+ C (n,n) == 2 ^ n,所以 cut = (2 ^ i - 1) % mod * (2 ^ (n - i) - 1) % mod

AC代码:

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
const int mod = 1e9 + 7;
const int K = 3e5 + 10;
int d[K],a[K];
LL C(int n){
    LL ans = 2,sum = 1;
    while(n){
        if(n & 1) sum = (sum * ans) % mod;
        ans = (ans * ans) % mod;
        n /= 2;
    }
    return (sum - 1 + mod) % mod;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
        scanf("%lld",&a[i]);
    sort(a + 1,a + 1 + n);
    for(int i = 1; i < n; i++)
        d[i] = a[i + 1] - a[i];
    LL ans = 0;
    for(int i = 1; i < n; i++)
        ans = (ans + C(i) * C(n - i) % mod * d[i] % mod) % mod;
    printf("%lld\n",ans);
    return 0;
}
### Codeforces 1732A Bestie 题目解析 对于给定的整数数组 \(a\) 和查询次数 \(q\),每次查询给出两个索引 \(l, r\),需要计算子数组 \([l,r]\) 的最大公约数(GCD)。如果 GCD 结果为 1,则返回 "YES";否则返回 "NO"[^4]。 #### 解决方案概述 为了高效解决这个问题,可以预先处理数据以便快速响应多个查询。具体方法如下: - **预处理阶段**:构建辅助结构来存储每一对可能区间的 GCD 值。 - **查询阶段**:利用已有的辅助结构,在常量时间内完成每个查询。 然而,考虑到内存限制以及效率问题,直接保存所有区间的结果并不现实。因此采用更优化的方法——稀疏表(Sparse Table),它允许 O(1) 时间内求任意连续子序列的最大值/最小值/GCD等问题,并且支持静态RMQ(Range Minimum Query)/RANGE_GCD等操作。 #### 实现细节 ##### 构建稀疏表 通过动态规划的方式填充二维表格 `st`,其中 `st[i][j]` 表示从位置 i 开始长度为 \(2^j\) 的子串的最大公约数值。初始化时只需考虑单元素情况即 j=0 的情形,之后逐步扩展至更大的范围直到覆盖整个输入序列。 ```cpp const int MAXN = 2e5 + 5; int st[MAXN][20]; // Sparse table for storing precomputed results. vector<int> nums; void build_sparse_table() { memset(st,-1,sizeof(st)); // Initialize the base case where interval length is one element only. for(int i = 0 ;i < nums.size(); ++i){ st[i][0]=nums[i]; } // Fill up sparse table using previously computed values. for (int j = 1;(1 << j)<=nums.size();++j){ for (int i = 0;i+(1<<j)-1<nums.size();++i){ if(i==0 || st[i][j-1]!=-1 && st[i+(1<<(j-1))][j-1]!=-1) st[i][j]=__gcd(st[i][j-1],st[i+(1<<(j-1))][j-1]); } } } ``` ##### 处理查询请求 当接收到具体的 l 和 r 参数后,可以通过查找对应的 log₂(r-l+1) 来定位合适的跳跃步长 k ,进而组合得到最终答案。 ```cpp string query(int L,int R){ int K=(int)(log2(R-L+1)); return __gcd(st[L][K],st[R-(1<<K)+1][K])==1?"YES":"NO"; } ``` 这种方法能在较短时间内完成大量查询任务的同时保持较低的空间开销,非常适合本题设定下的性能需求。
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