Triple Nim（Nim博弈）
Time Limit: 2000 ms Memory Limit: 65536 KiB
Submit Statistic Discuss
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
Multiple test cases. The first line contains an integer T (T <= 100000), indicating the number of test case. Each case contains one line, an integer N (3 <= N <= 1000000000) indicating the number of stones Bob have.
Output
One line per case. The number of ways Bob can make Alice never win.
Sample Input
3 3 6 14
Sample Output
0 1 4
Hint
In the third case, Bob can make three heaps (1,6,7), (2,5,7), (3,4,7) or (3,5,6).
Source
“浪潮杯”山东省第七届ACM大学生程序设计竞赛
题目简述：
给出n个石子，分成三堆， 对手先行，每次从一堆中拿走至少一块石子，拿走最后一堆最后一块的人胜利，求对手不可能胜利的方案数。
题目简析：
3堆石子的Nim博弈。要对手必败，则要使对手拿石子时3堆石子个数二进制异或值为0。
当N为奇数时自己必输，因为无论怎么分3堆石子的个数都会有两个是偶数，异或值不可能为0.
当N为偶数时，要保证总异或值为0，则每个二进制位要有两个1，所以要把给定的石头数N化成二进制，把每一位的1拆成另外两堆（把石子分成三堆，目前操作的假设为第一堆，其他两个分别为第二，第三堆）低一位的两个1，结束后二进制位全为0。然后对零进行排列，有3^num(num为把N化成二进制后1的个数)种排法，因为每一堆石子个数都不能为零所以要减3，得有3^num-3种排法，因为排列有重复如（1,2,3）和（3,2,1）为1种分法，所以要除6。
#include
#include
#include
typedef long long ll;
using namespace std;
int main()
{
int t,num;
ll ans;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
if(n%2) //奇数
printf("0\n");
else{
num = 0;
while(n){
if(n&1)
num++;
n>>=1;
}
ans = ((ll)pow(3,num)-3)/6;
printf("%lld\n",ans);
}
}
}