The area Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1637 Accepted Submission(s): 1298 Problem Description Ignatius bought a land last week, but he didn't know the area of the land because t
Problem Description There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2). Input Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000). Output Print the word "yes" if
题目链接:传送门 知识点: (1)三个点,三角形求面积公式 (2)精度问题: double 15-16位(参考文章) float 6-7位 long long 约20位 int 约10位 unsigned int 是int的两倍(参考文章) (3)nth_element()函数 思路:一开始想直接暴力求面积,然后面积排序,后来有发现面积不能是0,可以重复, 然后排序求出第k大的值,结果没注意double的位数不能精确到达到18位,然后又想四舍五入, 对尾数进行了处理,还是没过.看来题解才发现最后
今天推导公式,发现居然有对矩阵的求导,狂汗--完全不会.不过还好网上有人总结了.吼吼,赶紧搬过来收藏备份. 基本公式:Y = A * X --> DY/DX = A'Y = X * A --> DY/DX = AY = A' * X * B --> DY/DX = A * B'Y = A' * X' * B --> DY/DX = B * A' 1. 矩阵Y对标量x求导: 相当于每个元素求导数后转置一下,注意M×N矩阵求导后变成N×M了 Y = [y(ij)] --> dY/
Floating-Point Hazard 题目链接(点击) 题目描述 Given the value of low, high you will have to find the value of the following expression: If you try to find the value of the above expression in a straightforward way, the answer may be incorrect due to precision
题目链接:hdu_5810_Balls and Boxes 题意: 如题,让你求那个公式的期望 题解: 打表找规律,然后推公式.这项技能必须得学会 #include<cstdio> #include<algorithm> using namespace std; int main() { long long n,m,g; while(~scanf("%I64d%I64d",&n,&m)&&n+m) { n=n*(m-); m=m*
转载自: http://blog.csdn.net/txwh0820/article/details/46392293 矩阵的迹求导法则 1. 复杂矩阵问题求导方法:可以从小到大,从scalar到vector再到matrix 2. x is a column vector, A is a matrix d(A∗x)/dx=A d(xT∗A)/dxT=A d(xT∗A)/dx=AT d(xT∗A∗x)/dx=xT(AT+A) 3. Practice: 4. 矩阵求导计算法则 求导公式(撇号为