MAX Average Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7574 Accepted Submission(s): 1667
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].
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
6 4 2 10 3 8 5 9 4 1
Sample Output
6.50
题解:
http://www.docin.com/p-47950655.html这篇论文讲的斜率优化,讲的很清楚;
给定一个长度为n的序列,从其中找连续的长度大于m的子序列使得子序列中的平均值最小。
总结就是:构造下凸折线,维护下凸折线(凸包维护),找与下凸折线相切的斜率(也可以用二分来找)
没用二分:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define SD(x) scanf("%lf",&x)
#define P_ printf(" ")
typedef long long LL;
const int MAXN=1e5+;
int sum[MAXN],s[MAXN],a[MAXN];
bool cross(int i,int j,int k){
if((sum[j]-sum[i])*(k-i)>=(sum[k]-sum[i])*(j-i))return true;
return false;
}
double flx(int i,int t){
double temp;
temp=1.0*(sum[t]-sum[i])/(t-i);
return temp;
}
int main(){
int N,k;
while(~scanf("%d%d",&N,&k)){
sum[]=;
for(int i=;i<=N;i++)SI(a[i]),sum[i]=sum[i-]+a[i];
int top=,low=;
double ans=;
for(int i=k;i<=N;i++){
int j=i-k;
while(top-low>=&&cross(s[top-],s[top],j))top--;
s[++top]=j;
while(top-low>=&&flx(s[low+],i)>=flx(s[low],i))low++;
ans=max(ans,flx(s[low],i));
}
printf("%.2lf\n",ans);
}
return ;
}
用了二分:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define SD(x) scanf("%lf",&x)
#define P_ printf(" ")
typedef long long LL;
const int MAXN=1e5+;
int sum[MAXN],s[MAXN],a[MAXN];
bool cross(int i,int j,int k){
if((sum[j]-sum[i])*(k-i)>=(sum[k]-sum[i])*(j-i))return true;
return false;
}
double flx(int i,int t){
double temp;
temp=1.0*(sum[t]-sum[i])/(t-i);
return temp;
}
int erfen(int l,int r,int i){
int mid;
while(l<=r){
mid=(l+r)>>;
if(cross(s[mid],s[mid+],i))r=mid-;
else l=mid+;
}
return r+;
}
int main(){
int N,k;
while(~scanf("%d%d",&N,&k)){
sum[]=;
for(int i=;i<=N;i++)SI(a[i]),sum[i]=sum[i-]+a[i];
int top=,low=;
double ans=;
for(int i=k;i<=N;i++){
int j=i-k;
while(top-low>=&&cross(s[top-],s[top],j))top--;
s[++top]=j;
ans=max(ans,flx(s[erfen(,top,i)],i));
}
printf("%.2lf\n",ans);
}
return ;
}