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/* 威佐夫博奕（Wythoff Game） 两堆 对于这种博弈，有一个通项公式：a[k] = [k*(1+sqrt(5))/2],b[k] = a[k]+k; 基于两个规律.首先枚举前几个必败态 (0,0) (1,2) (3,5) (4,7) (6,10),.... 规律是，a[k] = 等于之前没有出现的自然数 b[k] = a[k]+k 通项公式的给出，还有一个定理，beatty定理： 正无理数，alpha,beta，如果有1/a+1/b=1,那么数列[an],[bn],严格递增，而且构成正整数上的一个划分 a[n] = [cn] b[n] = a[n]+n = [c*n] +n = [ 阅读全文
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/* 有关时针和分针夹角的问题*/// include file#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cctype>#include <ctime>#include <iostream>#include <sstream>#include <fstream>#include <iomanip>#include <bitset>#include & 阅读全文
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/* 当P>M的时候C有可能不等于P-M*/// include file#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cctype>#include <ctime>#include <iostream>#include <sstream>#include <fstream>#include <iomanip>#include <bitset>#i 阅读全文
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/* 不仅求最优解，还要过程,我的懒方法，执行两次。*/// include file#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cctype>#include <ctime>#include <iostream>#include <sstream>#include <fstream>#include <iomanip>#include <bitset> 阅读全文
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// include file#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cctype>#include <ctime>#include <iostream>#include <sstream>#include <fstream>#include <iomanip>#include <bitset>#include <algorithm># 阅读全文
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/* 不找规律，打表，搜索,O(n)*/// include file#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cctype>#include <ctime>#include <iostream>#include <sstream>#include <fstream>#include <iomanip>#include <bitset>#includ 阅读全文
摘要：
/* a[i] = (a[i-1]+a[i+1])/2-c[i]; (a[i]+c[i])*2=a[i-1]+a[i+1] a[i+1]= (a[i]+c[i])*2-a[i-1]; a[i] = (a[i-1]+c[i-1])*2-a[i-2] a[2] = (a[1]+c[1])*2-a[0] = 2*a[1]-a[0]+2*c[1] a[3] = (a[2]+c[2])*2-a[1] = ( (a[1]+c[1])*2-a[0]+c[2] )*2-a[1] = = 4*(a[1]+c[1])-2*a[0]+2*c[2]-a[1] = 3*a[1]-2*a[0]+2*c[2]+4*c[1] 阅读全文