思路
我们容易可以得到一个朴素的做法,首先对 \(a\) 数组排序,然后枚举最大值和最小值 \(a_i,a_j\),那么对于中间的元素都有选与不选两种情况,得到答案:
\[ \sum_{i = 1}^{n}(a_i \times a_i + (\sum_{j = i + 1}^{n}a_i \times a_j \times 2^{j - i - 1})) \]然后对这个式子做一个化简:
\[ \sum_{i = 1}^{n}(a_i \times a_i + a_i \times (\sum_{j = i + 1}^{n}a_j \times 2^{j - i - 1})) \]发现对于每一个 \(i\),\(a_j \times 2^{j - i - 1}\) 都是类似的,所以考虑预处理。
定义 \(m_i = \sum_{j = 1}^{i}(a_j \times 2^j)\),那么发现:
\[ m_n - m_i = \sum_{j = i + 1}^{n}{a_j}\times 2^j \]然后,发现对于每一项 \(j\) 对于原式都多乘了一个 \(2^{i + 1}\),直接除掉即可。得答案为:
\[ \sum_{i = 1}^n{(a_i \times a_i + \frac{m_n - m_i}{2^{i + 1}} \times a_i)} \]时间复杂度 \(\Theta(n \log n)\)。
Code
#include <bits/stdc++.h>
#define int long long
#define re register
using namespace std;
const int N = 2e5 + 10,mod = 998244353;
int n,ans;
int arr[N],pot[N],mul[N],inv[N];
inline int read(){
int r = 0,w = 1;
char c = getchar();
while (c < '0' || c > '9'){
if (c == '-') w = -1;
c = getchar();
}
while (c >= '0' && c <= '9'){
r = (r << 3) + (r << 1) + (c ^ 48);
c = getchar();
}
return r * w;
}
inline int Add(int a,int b){
return (a + b) % mod;
}
inline int Sub(int a,int b){
return ((a - b) % mod + mod) % mod;
}
inline int Mul(int a,int b){
return a * b % mod;
}
inline void exgcd(int a,int b,int &x,int &y){
if (!b){
x = 1;
y = 0;
return;
}
exgcd(b,a % b,y,x);
y = y - a / b * x;
}
inline int get_inv(int a,int p){
int x,y;
exgcd(a,p,x,y);
return (x % mod + mod) % mod;
}
inline void init(){
pot[0] = 1;
for (re int i = 1;i <= n + 1;i++){
pot[i] = Mul(pot[i - 1],2);
mul[i] = Add(mul[i - 1],Mul(arr[i],pot[i]));
inv[i] = get_inv(pot[i],mod);
}
}
signed main(){
n = read();
for (re int i = 1;i <= n;i++) arr[i] = read();
sort(arr + 1,arr + n + 1);
init();
for (re int i = 1;i <= n;i++){
ans = Add(ans,Mul(Mul(Sub(mul[n],mul[i]),inv[i + 1]),arr[i]));
ans = Add(ans,Mul(arr[i],arr[i]));
}
printf("%lld",ans);
return 0;
}
标签:ARC116B,arc116,Min,int,题解,sum,times
From: https://www.cnblogs.com/WaterSun/p/18263293