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取(2堆)石子游戏

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2054    Accepted Submission(s): 1238

1 2 
5 8
4 7
2 2
0 0

0
1
4 7
3 5
0
1
0 0
1 2

题意：中文题.....不说话.......

分析：题是基础题，但是要懂具体是怎么拿的，不懂的话.....就暴力嘛，这数据也不大嘛.....

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 1e9;
const int MOD = 1e9+7;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define lson (i<<1)
#define rson ((i<<1)|1)
#define N 50010
int gcd(int a,int b){return b?gcd(b,a%b):a;}

int main()
{
    int n,m,a,b,k;
    double x = (1 + sqrt(5.0))/2.0;
    while(scanf("%d%d",&a,&b),(a+b))
    {
        if(a>b) swap(a, b);
        k = b-a;
        if(a==(int)(k*x)) cout<<0<<endl;
        else
        {
            cout<<1<<endl;
            for(int i=1; i<=a; i++)
            {
                n = a-i;
                m = b-i;
                k = m-n;
                if(n==(int)(k*x)) cout<<n<<" "<<m<<endl;
            }
            for(int i=b-1; i>=0; i--)
            {
                n = a;
                m = i;
                if(n>m) swap(n, m);
                k = m-n;
                if(n==(int)(k*x)) cout<<n<<" "<<m<<endl;
            }
        }
    }
    return 0;
}

hdu2177 取(2堆)石子游戏 （威佐夫博弈）

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原文地址：http://blog.csdn.net/d_x_d/article/details/52124601