sum[i]为前缀和
对于以i结尾的一段满足要求的最大和则为x=sum[i]-sum[j-1],i-j+1<=k;
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef pair<int,int> pii; #define pb(a) push_back(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c){return min(min(a,b),min(a,c));} template<class T> T max(const T& a,const T& b,const T& c){return max(max(a,b),max(a,c));} void debug() { #ifdef ONLINE_JUDGE #else freopen("d:\in.txt","r",stdin); //freopen("d:\out1.txt","w",stdout); #endif } char getch() { char ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!=' ')return ch; } return EOF; } int da[100100],sum[200200],q[200200]; int main() { debug(); int t; scanf("%d",&t); for(int ca=1;ca<=t;ca++) { int n,k; scanf("%d%d",&n,&k); sum[0]=0; for(int i=1;i<=n;i++) { scanf("%d",&da[i]); sum[i]=sum[i-1]+da[i]; } for(int i=1;i<=n;i++) { sum[i+n]=sum[i+n-1]+da[i]; } int front=1,rear=1; q[1]=0; int maxx=INT_MIN,ml,mr; for(int i=1;i<=2*n;i++) { while(front<=rear&&i-q[front]>k)front++; if(sum[i]-sum[q[front]]>maxx) { maxx=sum[i]-sum[q[front]]; ml=q[front]+1;mr=i; } while(front<=rear&&sum[i]<sum[q[rear]])rear--; q[++rear]=i; } if(mr>n)mr=mr%n; printf("%d %d %d ",maxx,ml,mr); } return 0; }