Topcoder SRM 547 DIV1 Level 1: Pillars

题目： http://community.topcoder.com/stat?c=problem_statement&pm=12055&rd=14739
题目大意：对所有的1<=x<=X， 1<=y<=Y和给定的w， 求s = sqrt（w*w + （x-y）*（x-y））的平均值。

解题报告：

如果枚举所有可能情况，则时间复杂度为O（X* Y），该题中X和Y分别最大值为100000，显然暴力枚举过不了测试。具体分析发现公式S显然只随着d=y-x在变化，原来的枚举空间中有很多次重复的计算。假设y>x（如果不大于，交换两个数）， 则d=1,2,...,K=Y-X 分别有X条；d= K+1， K+2, ..., Y-1分别有X-1,X-2,...,1条； d= -1, -2, ..., -X+1分别有X-1,X-2,...,1条。分别求出d=1，2，。。。，Y-1的各个边长，再求和取平均，即可得最终结果。
时间复杂度为O（max{X,Y }）.

 

解法二：动态规划求解

Let's have a look at the sum of all possible  lengthes. Let d(i, j) be the length between two pillars with heights i and j. Then we have the following sum:

S=d(1,1)+d(1,2)+...+d(1,y)+d(2,1)+d(2,2)+...+d(2,y)+...+d(x,1)+...+d(x,y).

Obviuosly d(i,j)=d(j,i) and d(i-1,j-1)=d(i,j). We can assume that x<=y and  then:

S=d(1,1)+d(1,2)+...+d(1,y)+d(1,2)+d(1,1)+d(1,2)+..+d(1,y-1)+...+d(1,x)+d(1,x-1)+...+d(1,y-x+1).

Let f(n) be the sum d(1,2)+...+d(1,n). Then

S=f(y)+f(1)+d(2,2)+f(y-1)+f(2)+d(3,3)+f(y-2)+...+f(x-1)+d(x,x)+f(y-x+1).

s = sum {f[x] + w + f[y-x+1]} {1<=x<=X, w = d[x,x]}

 

Code

class Pillars
{

public double getExpectedLength(int w, int x, int y)
{
double result = 0;
if (y < x)
{
int tmp = y;
y = x;
x = tmp;
}
int delta = y - x;
for (int i = 0; i <= delta; i++)
{
double z = 1.0 * x;
result += z * System.Math.Sqrt( 1.0 *w * w + 1.0 * i * i);
}

for (int i = delta + 1; i < y; i++)
{
double z = 1.0 * (y - i);
result += z * System.Math.Sqrt(1.0 * w * w + 1.0 * i * i);
}

for (int i = 1; i < x; i++)
{
double z = 1.0 * (x - i);
result += z * System.Math.Sqrt(1.0 * w * w + 1.0 * i * i);
}
return result/x/y;
}

public double getExpectedLength2(int w, int x, int y)
{
double[] f = new double[100001];
int N = System.Math.Max(x, y);
f[1] = 0;
for (int i = 2; i <= N; i++)
{
double d = distance(w, i - 1);
f[i] = (f[i - 1] + d);
}
double sum = 0;
for (int i = 1; i <= x; i++)
{
sum += f[i] + w + f[y - i + 1];
}

return sum / x / y;
}

private static double distance(int w, int d)
{
double x = 1.0 * w * w + 1.0 * d * d;
return System.Math.Sqrt(x);
}
}