标签：.sh The value AMM val pre review img integer
B Hanoi tower
It has become a good tradition to solve the “Hanoi tower” puzzle at programming contests in Rybinsk. We will review the rules briefly.
2*n/3-1移到C上，再从A移动一个到B上
再把C的n/3-1移到B上，现在所有上面都是1/3了
队友得到了一个神奇的公式2^(n-n/3-1)+2^(n/3-1)-1
但是经过提交是不行的，所以暴力打表找规律
#include <stdio.h> using namespace std; int num[3],f,f1; int move(int n,int a,int b) { //printf("Move disk %d from %c to %c\n",n,a,b); num[a]--,num[b]++; if(num[0]==num[1]&&num[1]==num[2])return 1; return 0; } void hanoi(int n,int a,int b,int c) { if(f1)return; if(n==1) { if(move(n,a,c)) { printf("%d\n",f); f1=1; return; } } else { hanoi(n-1,a,c,b); f++; if(move(n,a,c)) { printf("%d\n",f); f1=1; return; } hanoi(n-1,b,a,c); } } int main() { int n; for(int n=3; n<=30; n+=3) { num[0]=n,num[1]=num[2]=f=0,f1=0; hanoi(n,0,1,2); } return 0; }
很快就会发现偶数的猜想是对的，所以对奇数进行讨论，发现正好是*4+2
暴力代码，交的表，因为莫名RE
import java.math.*; import java.util.*; public class Main { public static void main(String args[]) { Scanner cin=new Scanner(System.in); BigInteger tmp=BigInteger.ZERO; for(int n=3; n<=300; n+=3) { if(n/3%2==1) { System.out.println("\""+tmp.multiply(BigInteger.valueOf(4)) .add(BigInteger.valueOf(2))+"\","); } else { int x=n-n/3-1; int y=n/3-1; tmp=cal(x).add(cal(y)).subtract(BigInteger.ONE); System.out.println("\""+tmp+"\","); } } } static BigInteger cal(int x) { BigInteger ans=BigInteger.ONE; ans=ans.shiftLeft(x); return ans; } }
2016-2017 ACM Central Region of Russia Quarterfinal Programming Contest BHanoi tower
标签：.sh The value AMM val pre review img integer
原文地址：https://www.cnblogs.com/BobHuang/p/9664780.html
