Triple Nim
Problem Description
Alice and Bob are always playing all kinds of Nim games and Alice always goes first. Here is the rule of Nim game:
There are some distinct heaps of stones. On each turn, two players should remove at least one stone from just one heap. Two player will remove stone one after another. The player who remove the last stone of the last heap will win.
Alice always wins and Bob is very unhappy. So he decides to make a game which Alice will never win. He begins a game called “Triple Nim”, which is the Nim game with three heaps of stones. He’s good at Nim game but bad as math. With exactly N stones, how many ways can he finish his target? Both Alice and Bob will play optimally.
Input
Output
Example Input
33614
Example Output
014
Hint
Author
题意:给你n个石子,分成三堆,计算通过Nim博弈的规则使得对方不能获胜的方案数。
思路:打表找规律题,不解释。
参考网址:http://blog.csdn.net/huayunhualuo/article/details/51626212
#include <bits/stdc++.h> using namespace std; typedef long long LL; int main() { int T; LL n; scanf("%d",&T); while(T--) { scanf("%lld",&n); if(n%2) printf("0 "); else { int s = 0 ; while(n) { if(n%2) s++; n/=2; } LL ans = (LL(pow(3,s))-3)/6; printf("%lld ",ans); } } return 0; }