ArcTanWhen the ArcTan functional configuration is selected, the input vector (X_IN,Y_IN) is rotated(using the CORDIC algorithm) until the Y component is zero. This generates the outputangle, Atan(Y_IN/X_IN).The inputs, X_IN and Y_IN, are limited to t…
POJ1183 除输入方式外与这道题完全一样 题目大意是给定一个a 求最小的满足arctan(1/A)=arctan(1/B)+arctan(1/C) 的B+C的最小值 根据上述递推规律,我们只要从2*a开始往前递增寻找找到第一个满足b,c均为正整数的情况就是最小的 #include <cstdio> #define ll long long int main() { int T,n; scanf("%d",&T); while(T--){ scanf("…
Use of Function Arctan Time Limit:10000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Status Description It's easy to know that arctan(1/2)+arctan(1/3)=arctan(1).The problem is,to some fixed number A,you have to write a program to c…
\[\Large\int_0^1\frac{\arctan x \,\operatorname{arctanh} x\, \ln x}{x}\mathrm{d}x=\frac{\pi^2}{16}\mathbf{G}-\frac{7\pi}{32}\zeta(3)\] \(\Large \mathrm{\mathbf{Proof:}}\) Let \(n=0,1,2,\cdots\), We define \(I,I_{1,n},I_{2,n}\) and \(I_n\) as follows:…
详细的题解见这里. 图片转自上面的博客 假设我们已经推导出来x在处取得最小值,并且注意到这个点是位于两个整点之间的,所以从这两个整数往左右两边枚举b就能找到b+c的最小值. 其实只用往一边枚举就够了,由于对称性,我们不妨假设b ≤ c,那么只要让b从2a开始递减枚举即可. #include <iostream> using namespace std; long long a, i; ;)%(i-a))i--;cout<<(i*i+)/(i-a)<<endl;};} 代…