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原题地址:http://poj.org/problem?id=2018
题意: 给你N个数，让你求一个连续数串的平均值最大，连续数串的个数不得低于F个。
思路:二分平均值,然后对于每一个平均值,去进行判断是否存在一段子序列长度大于L,且平均值最大.然后可以对每一个数先减去平均值,然后利用类似于双指针的思想去求解.
注意:这题好像有点坑,没有spj,因此不能二分循环100次.
ac代码:
#include <cmath>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <set>
#include <map>
#include <cctype>
#define eps 1e-5
#define INF 1e12
#define PI acos(-1)
#define lson l,mid,rt<<1
#define rson mid+1,r,(rt<<1)+1
#define CLR(x,y) memset((x),y,sizeof(x))
#define fuck(x) cerr << #x << "=" << x << endl
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int seed = 131;
const int maxn = 1e5 + 5;
const int mod = 998244353;
int n, f;
double a[maxn], b[maxn], sum[maxn];
bool check(double mid) {
for (int i = 1; i <= n; i++) {
b[i] = a[i] - mid;
sum[i] = sum[i - 1] + b[i];
}
double min_val = INF;
for (int i = f; i <= n; i++) {
min_val = min(min_val, sum[i - f]);
if (sum[i] - min_val >= 0) return 1;
}
return 0;
}
int main() {
scanf("%d%d", &n, &f);
for (int i = 1; i <= n; i++) scanf("%lf", &a[i]);
double l = 0;
double r = INF;
while (r - l > eps) {
double mid = (l + r) / 2;
if (check(mid)) l = mid;
else r = mid;
}
printf("%d\n", (int) (r * 1000));
return 0;
}