Poj2018 best cow fences

My understanding of binary: https://www.cnblogs.com/AKMer/p/9737477.html

Question Portal: http://poj.org/problem? Id = 2018

Let's divide an average into two parts, and set \ (A \) array to each number minus the average to \ (B \) array. If \ (B \) an array with a length greater than \ (k \) and a weight greater than \ (0 \) indicates that the final average value must be greater than the average value of our current binary.

So how can we find the weight and length of an array greater than \ (0 \) and greater than \ (k? Obviously, we can enumerate the following fields:

for(int i=1;i<=n;i++)    for(int j=0;j<=i-k;j++)        if(sum[i]-sum[j]>0)return 1;

But this \ (N ^ 2 \) method obviously does not work. We can find that the Section ending with \ (I \) is \ (sum [0] ~ The minimum value in sum [I-K] \) is \ (sum [I]-Mn> 0. So we only need to \ (I + 1 \), \ (Mn \) and \ (sum [I-k + 1] \) take \ (Min \) that's all.

Time Complexity: \ (O (nloga )\)

Space complexity: \ (O (n )\)

The Code is as follows:

# Include <cstdio> # include <algorithm> using namespace STD; const double EPS = 1e-6; const int maxn = 1e5 + 5; int N, K; int A [maxn]; double sum [maxn]; int read () {int x = 0, F = 1; char CH = getchar (); (; ch <'0' | ch> '9'; CH = getchar () if (CH = '-') F =-1; (; CH> = '0' & Ch <= '9'; CH = getchar () x = x * 10 + CH-'0'; return x * F ;} bool check (double AVE) {for (INT I = 1; I <= N; I ++) sum [I] = sum [I-1] + (double) A [I]-Ave; double Mn = 0; For (INT I = K; I <= N; I ++) {If (sum [I]-Mn> 0) return 1; Mn = min (Mn, sum [I-k + 1]);} return 0;} int main () {double L = 1e9, r =-1e9; N = read (), k = read (); For (INT I = 1; I <= N; I ++) {A [I] = read (); L = min (L, (double) A [I]); r = max (R, (double) A [I]);} while (L + EPS <R) {double mid = (L + r)/2; If (check (MID) L = mid; else r = mid;} printf ("% d \ n ", (INT) (R * 1000); // output R * 1000 return 0 because the average value must be greater than l ;}

Poj2018 best cow fences