Maximize The Beautiful Value
网址:牛客网
题目大意
- 给定一个序列,序列价值为:a[i] * i
- 有且仅有一次交换,将一个值提前最少k个,其他相对顺序不变
- 求序列价值
解题思路
使用前缀和找到最小差值相减
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep2(i,st,ed) for(int i = st; i < ed; ++i)
#define mk(x,y) make_pair(x,y)
#define pb(x) push_back(x)
const ll mod = 1e9 + 7;
const ll N = 2e5 + 200;
const ll INF = 1e18;
const double eps = 1e-7;
int w[N], v[N];
ll dp[N];
int main(){
int t;
cin>>t;
while(t--){
ll n,k;
cin>>n>>k;
vector<ll> a(n+1, 0);
vector<ll> b(n+1, 0);
ll sum = 0;
rep2(i,1,n+1){
cin>>a[i];
sum += i * a[i];
b[i] = b[i-1] + a[i];
}
ll mi = -INF;
// cout<<sum<<endl;
rep2(i,k,n+1){
ll tmp = b[i-1] - b[i-k-1] - a[i] * k;
mi = max(mi, tmp);
}
sum += mi;
cout<<sum<<endl;
}
return 0;
}