MAX Average Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5825 Accepted Submission(s): 1446
Problem Description
Consider a simple sequence which only contains positive integers as a1, a2 ... an, and a number k. Define ave(i,j) as the average value of the sub sequence ai ... aj, i<=j. Let’s calculate max(ave(i,j)), 1<=i<=j-k+1<=n.
Input
There multiple test cases in the input, each test case contains two lines.
The first line has two integers, N and k (k<=N<=10^5).
The second line has N integers, a1, a2 ... an. All numbers are ranged in [1, 2000].
The first line has two integers, N and k (k<=N<=10^5).
The second line has N integers, a1, a2 ... an. All numbers are ranged in [1, 2000].
Output
For every test case, output one single line contains a real number, which is mentioned in the description, accurate to 0.01.
Sample Input
10 6 6 4 2 10 3 8 5 9 4 1
Sample Output
6.50
直接斜率DP:O(N)
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <cmath>
#include <map>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;
const int MAX=100000+10;
int n,k;
int s[MAX],q[MAX];
double dp[MAX],sum[MAX];
double GetY(int i,int j){
return sum[i]-sum[j];
}
int GetX(int i,int j){
return i-j;
}
double DP(){
int head=0,tail=1;
q[head]=0;
double ans=0;
for(int i=1;i<=n;++i)sum[i]=sum[i-1]+s[i]*1.0;
for(int i=k;i<=n;++i){
int j=i-k;
while(head+1<tail && GetY(j,q[tail-1])*GetX(q[tail-1],q[tail-2])<=GetY(q[tail-1],q[tail-2])*GetX(j,q[tail-1]))--tail;
q[tail++]=j;
while(head+1<tail && GetY(i,q[head])*GetX(i,q[head+1])<=GetY(i,q[head+1])*GetX(i,q[head]))++head;
dp[i]=(sum[i]-sum[q[head]])/(i-q[head]);
ans=max(ans,dp[i]);
}
return ans;
}
int input(){//加速外挂
char ch=' ';
int num=0;
while(ch<'0' || ch>'9')ch=getchar();
while(ch>='0' && ch<='9')num=num*10+ch-'0',ch=getchar();
return num;
}
int main(){
while(~scanf("%d%d",&n,&k)){
for(int i=1;i<=n;++i)s[i]=input();
printf("%0.2lf
",DP());
}
return 0;
}
斜率DP+二分查找:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <cmath>
#include <map>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;
const int MAX=100000+10;
int n,k;
int s[MAX],q[MAX];
LL sum[MAX];
LL GetY(int i,int j){
return sum[i]-sum[j];
}
int GetX(int i,int j){
return i-j;
}
LL check(int mid,int i){
return GetY(i,q[mid+1])*GetX(q[mid+1],q[mid])-GetY(q[mid+1],q[mid])*GetX(i,q[mid+1]);
}
int search(int l,int r,int i){
//由于斜率单调递增
/*int top=r;
while(l<=r){//依据i与mid的斜率 和 i与mid+1的斜率之差求切点
if(l == r && l == top)return q[l];//这里一定要注意假设切点是最后一个点须要另判,由于mid+1不存在会出错
int mid=(l+r)>>1;
if(check(mid,i)<0)r=mid-1;
else l=mid+1;
}*/
while(l<r){//依据i与mid的斜率 和 i与mid+1的斜率之差求切点
int mid=(l+r)>>1;
if(check(mid,i)<0)r=mid;
else l=mid+1;
}
return q[l];
}
double DP(){
int head=0,tail=1,p;
q[head]=0;
double ans=0,dp;
for(int i=1;i<=n;++i)sum[i]=sum[i-1]+s[i];
for(int i=k;i<=n;++i){
int j=i-k;
while(head+1<tail && GetY(j,q[tail-1])*GetX(q[tail-1],q[tail-2])<=GetY(q[tail-1],q[tail-2])*GetX(j,q[tail-1]))--tail;
q[tail++]=j;
p=search(head,tail-1,i);//依据相邻点与i点的斜率之差二分查找切点
dp=(sum[i]-sum[p])*1.0/(i-p);
if(dp>ans)ans=dp;
}
return ans;
}
int input(){//加速外挂
char ch=' ';
int num=0;
while(ch<'0' || ch>'9')ch=getchar();
while(ch>='0' && ch<='9')num=num*10+ch-'0',ch=getchar();
return num;
}
int main(){
while(~scanf("%d%d",&n,&k)){
for(int i=1;i<=n;++i)s[i]=input();
printf("%0.2lf
",DP());
}
return 0;
}