首页 > 代码库 > HDU 2993 MAX Average Problem（斜率优化）

HDU 2993 MAX Average Problem（斜率优化）

2024-07-19 17:27:48   217人阅读

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2993

Problem Description
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.
 
Input
There multiple test cases in the input, each test case contains two lines.
The first line has two integers, N and k (k<=N<=10^5).
The second line has N integers, a1, a2 ... an. All numbers are ranged in [1, 2000].
 
Output
For every test case, output one single line contains a real number, which is mentioned in the description, accurate to 0.01.
 
题目大意：给n个数和k，求数的个数大于等于k的子段的最大平均值。
思路：可以去看IOI国家集训队论文：《浅谈数形结合思想在信息学竞赛中的应用》——周源
也可以直接看这个http://blog.sina.com.cn/s/blog_ad1f8960010174el.html
 
PS：这就是传说中的来自数据组数的恶意吗，看上去似乎有100组大数据的感觉……G++超时的可以尝试用C++交，HDU的C++读入比G++快，而且优化的程度也不同。
 
代码（C++ 500MS/G++ 906MS）：
 1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <cctype> 6 using namespace std; 7 typedef long long LL; 8  9 const int MAXN = 100010;10 11 int sum[MAXN];12 int n, k;13 14 int readint() {15     char c = getchar();16     while(!isdigit(c)) c = getchar();17     int res = 0;18     while(isdigit(c)) res = res * 10 + c - ‘0‘, c = getchar();19     return res;20 }21 22 struct Point {23     int x, y;24     Point() {}25     Point(int x, int y): x(x), y(y) {}26     Point operator - (const Point &rhs) const {27         return Point(x - rhs.x, y - rhs.y);28     }29     double slope() {30         return (double)y / x;31     }32 };33 34 LL cross(const Point &a, const Point &b) {35     return (LL)a.x * b.y - (LL)a.y * b.x;36 }37 38 LL cross(const Point &o, const Point &a, const Point &b) {39     return cross(a - o, b - o);40 }41 42 Point que[MAXN], p;43 int head, tail;44 45 double solve() {46     double res = 0;47     head = 0; tail = -1;48     for(int i = k; i <= n; ++i) {49         p = Point(i - k, sum[i - k]);50         while(head < tail && cross(que[tail - 1], que[tail], p) <= 0) --tail;51         que[++tail] = p;52 53         p = Point(i, sum[i]);54         while(head < tail && cross(que[head], que[head + 1], p) >= 0) ++head;55         res = max(res, (p - que[head]).slope());56     }57     return res;58 }59 60 int main() {61     while(scanf("%d%d", &n, &k) != EOF) {62         for(int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + readint();63         printf("%.2f\n", solve());64     }65 }
View Code

 

HDU 2993 MAX Average Problem（斜率优化）


联系
我们