problem
给定一个正整数数列A,求一个平均数最大、长度不小于L的子段。
solution
二分判定:是否存在一个长度大于L,平均数不小于二分值的子段。
codes
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
double a[100001], b[100001], sum[100001];
int main(){
int N, L;
cin >> N >> L;
for(int i = 1; i <= N; i++) scanf("%lf", &a[i]);
double eps = 1e-5;
double l = -1e6, r = 1e6;
while(r - l > eps){
double mid = (l + r) / 2;
for (int i = 1; i <= N; i++) b[i] = a[i] - mid;
for (int i = 1; i <= N; i++)
sum[i] = (sum[i - 1] + b[i]);
double ans = -1e10;
double min_val = 1e10;
for (int i = L; i <= N; i++) {
min_val = min(min_val, sum[i - L]);
ans = max(ans, sum[i] - min_val);
}
if (ans >= 0) l = mid; else r = mid;
}
cout << int(r * 1000) << endl;
}