第七届ACM山东省赛-G Triple Nim

Time Limit: 2000MS Memory limit: 65536K

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.

输入
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.
输出
One line per case. The number of ways Bob can make Alice never win.

示例输入

3
3
6
14

示例输出

0
1
4

提示

In the third case, Bob can make three heaps (1,6,7), (2,5,7), (3,4,7) or (3,5,6).

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。

#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\n");
        else 
        {
            int s = 0 ;
            while(n)
            {
                if(n%2) s++;

                n/=2;
            }

            LL ans = (LL(pow(3,s))-3)/6;

            printf("%lld\n",ans);
        }
    }
    return 0;
}