A Simple Stone Game Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0 Problem Description After he has learned how to play Nim game, Bob begins to try another ston
题目链接: GTW likes function Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Problem Description Now you are given two definitions as follows. f(x)=∑xk=0(−1)k22x−2kCk2x−k+1,f0(x)=f(x),fn(x)=f(fn−1(x))(n≥1) Note th
http://blog.sina.com.cn/s/blog_6f3a860501019z1f.html #include<iostream> #include<algorithm> using namespace std; int main() { int n[]={1,4,22,3,8,5}; int len=sizeof(n)/sizeof(int); cout<<*max_element(n,n+len)<<endl; cout<<*mi
octave:14> help plot'plot' is a function from the file C:\Octave\Octave3.6.4_gcc4.6.2\share\octave\3.6.4\m\plot\plot.m -- Function File: plot (Y) -- Function File: plot (X, Y) -- Function File: plot (X, Y, PROPERTY, VALUE, ...) -- Function File: plot
把一对石子堆看出一个子游戏.打出子游戏的sg表找规律.. 这个规律我是一定找不出来的... 对于i,j,如果 (i-1)%pow(2,k+1) < pow(2,k) (j-1)%pow(2,k+1) < pow(2,k) 那么最小的k值就是sg值. # include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vect
#include <stdio.h> int fun(int x) { int a, b, c; a = x / ; b = x % / ; c = x % ; if (x == a * a * a + b * b * b + c * c * c) ; else ; } int main() { int m; printf("1000以内的水仙花数:\n"); ; m < ; m++) { ) printf("%5d\n", m); } }