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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