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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6277 Accepted Submission(s): 2289
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
using namespace std;
#define INF 100000000
int tt,n,k;
int sum[200005],a[200005],hd,ed;
int main()
{
scanf("%d",&tt);
while(tt--)
{
scanf("%d%d",&n,&k);
int j=n;
sum[0]=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum[i]=sum[i-1]+a[i];
}
for(int i=n+1;i<n+k;i++)
{
sum[i]=sum[i-1]+a[i-n];
}
n=n+k-1;
int ans=-INF;
deque<int> q;
q.clear();
for(int i=1;i<=n;i++)
{
while(!q.empty()&&sum[i-1]<sum[q.back()])//维护单调性
q.pop_back();
while(!q.empty()&&q.front()<i-k)
q.pop_front();
q.push_back(i-1);
if(sum[i]-sum[q.front()]>ans)
{
ans=sum[i]-sum[q.front()];
hd=q.front()+1;
ed=i;
}
}
if(ed>j)
ed=ed%j;
printf("%d %d %d\n",ans,hd,ed);
}
return 0;
}
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原文地址:http://www.cnblogs.com/a972290869/p/4419159.html