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].
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 ", solve()); 64 } 65 }