最好的牛栏 POJ2018
思路
题干在这:POJ2018
本题就是求一个正整数数列中平均数最大且长度不小于 f 的子段
二分最大的平均数,让每个元素都减去这个平均数,再求子段和,如果最大的子段和大于0,就说明平均数还能更大,以此二分答案。注意输出要求,二分左右端点l,r应该输出1000*r(结果应该是一个整数的话,l * 1000会少一)
ac代码
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<math.h>
#define eps 1e-6
#define MAX 100001
using namespace std;
int number[MAX] = { 0 };
double sum[MAX] = { 0 };
int main() {
int n, f;
cin >> n >> f;
int maxi = 0;
for (int i = 1; i <= n; i++) {
cin >> number[i];
maxi = max(number[i], maxi);
}
double l = 0, r = maxi;
double average;
while (fabs(r - l) > eps) {
average = (1.0*l + 1.0*r) / 2;
memset(sum, 0, sizeof(sum));
for (int i = 1; i <= n; i++) {
sum[i] = sum[i - 1] + number[i] - average;
}
double maxaverage = sum[f], minsum = 0;
for (int i = f + 1; i <= n; i++) {
minsum = min(minsum, sum[i - f]);
maxaverage = max(maxaverage, sum[i] - minsum);
}
if (maxaverage >= 0)
l = average;
else
r = average;
}
int resort = int(r * 1000);
cout << resort;
}