题目链接:https://www.luogu.com.cn/problem/P5638
- 前缀和模拟
#include<bits/stdc++.h>
using namespace std;
const int max_n=1e6+10;
int n, k;
long long a[max_n];
long long s[max_n];
long long ans;
int main()
{
cin>>n>>k;
for(int i=2; i<=n; i++){//注意此处为n-1段距离
cin>>a[i];
s[i]=s[i-1]+a[i];
}
long long maxlen=0;
for(int i=2; i<=n; i++){
if(k==0)//k为0时找最大距离值
maxlen=max(maxlen, a[i]);
else if(i+k-1 <= n && s[i+k-1]-s[i-1] > maxlen)//最大区间和,依次枚举+k-1处,无需枚举-k处
maxlen=s[i+k-1]-s[i-1];
}
ans=s[n]-maxlen;
cout<<ans;
return 0;
}
标签:int,max,CSGRound2,荣耀,long,maxlen,P5638,ans
From: https://www.cnblogs.com/tflsnoi/p/17146678.html