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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.
There multiple test cases in the input, each test case contains two lines.
The first line has two integers, N and k (k<=M<=10^5).
The second line has N integers, a1, a2 ... an. All numbers are ranged in [1, 2000].
For every test case, output one single line contains a real number, which is mentioned in the description, accurate to 0.01.
10 6 6 4 2 10 3 8 5 9 4 1
6.50
1 #include <stdio.h> 2 #include <iostream> 3 using namespace std; 4 typedef long long LL; 5 const int maxn = 100005; 6 int q[maxn],hd,tl,n,k; 7 LL sum[maxn]; 8 bool check(LL a,LL b,LL c){ 9 return (sum[c] - sum[b])*(b - a) <= (sum[b] - sum[a])*(c - b); 10 } 11 int main(){ 12 while(~scanf("%d%d",&n,&k)){ 13 for(int i = 1; i <= n; ++i){ 14 scanf("%I64d",sum + i); 15 sum[i] += sum[i-1]; 16 } 17 double ret = 0; 18 hd = tl = 0; 19 for(int i = k; i <= n; ++i){ 20 while(hd + 1 < tl && check(q[tl-2],q[tl-1],i - k)) --tl; 21 q[tl++] = i - k; 22 while(hd + 1 < tl && check(q[hd+1],q[hd],i)) ++hd; 23 ret = max(ret,(sum[i] - sum[q[hd]])*1.0/(i - q[hd])); 24 } 25 printf("%.2f\n",ret); 26 } 27 return 0; 28 }
BNUOJ 3958 MAX Average Problem
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原文地址：http://www.cnblogs.com/crackpotisback/p/4803572.html
