题目传送门
威佐夫博弈传送门
思路:由于题目数据范围非常大,需要做高精度处理,难点在于计算(n-m)*(sqrt(5)+1)/2的整数部分值,可以采用特殊办法预处理(sqrt(5)+1)/2的值,再做大数乘法,题目难度会小很多,matlab有非常强大的高精度计算功能,在命令行输入 :
vpa((1 + sqrt(sym(5)))/2,100)
即可计算(sqrt(5)+1)/2精确到小数点后100位。
完整代码如下:
#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
using namespace std;
typedef unsigned long ul;
string n,m;
//if a>=b return true;
bool compare(string a,string b)
{
if(a[0] == '-')
return false;
if(a.size()>b.size())
return true;
else if(a.size()==b.size())
return a>=b;
return false;
}
string add(string,string);
string sub(string,string);
string mul(string,int);
string mul(string,string);
bool Wythoff(int n,int m)
{
if(n<m) swap(n,m);
int k=n-m;
n=(int)(k*(1+sqrt(5))/2.0);
if(n==m)
return 0;
else
return 1;
}
int main()
{
string s("61803398874989484820458683436563811772030917980576286213544862270526046281890244970720720418939113748");
while(cin>>n>>m)
{
if(n=="0"&&m=="0")
{
cout<<0<<endl;
continue;
}
if(compare(m, n))
swap(n, m);
string k = sub(n, m);
string ss(101-k.length(),'0');
string a(k);
a+=ss;
a = mul(a, s);
string cnt;
cnt = a.substr(0,a.length()+k.length()-202);
k = add(k , cnt);
if(k==m)
cout<<0<<endl;
else
cout<<1<<endl;
}
return 0;
}
string add(string a,string b)
{
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
if(a.size()<b.size())
swap(a, b);
bool f = 0;
for(int i=0;i<b.size();i++)
{
a[i] = a[i] +b[i]-48+f;
if(a[i]>'9')
{
f = true;
a[i] = a[i] -10;
}
else
f = false;
}
for(ul i=b.size();i<a.size();i++)
{
a[i] = a[i]+f;
if(a[i]>'9')
{
f = true;
a[i] = a[i] -10;
}
else
f = false;
}
if(f)
a = a+"1";
reverse(a.begin(),a.end());
return a;
}
string sub(string a,string b)
{
bool flag = compare(a, b);
if(!flag)
swap(a,b);
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
bool f = false;
for(int i=0;i<b.size();i++)
{
a[i] = a[i] + '0' - b[i] -f;
if(a[i]<'0')
{
f = true;
a[i] = a[i] + 10;
}
else
f = false;
}
for(ul i=b.size();i<a.size();i++)
{
a[i] = a[i] - f;
if(a[i]<'0')
{
f = true;
a[i] = a[i] + 10;
}
else
f = false;
}
reverse(a.begin(),a.end());
while(*a.begin()=='0')
a.erase(a.begin());
if(a.size()==0)
a = "0";
if(!flag)
a = "-"+a;
return a;
}
string mul(string a,int b)
{
int flag = 0;
for(int i=a.size()-1;i>=0;i--)
{
int str = (a[i]-48)*b + flag;
a[i] = str%10 +48;
flag = str/10;
}
if(flag)
a.insert(a.begin(), flag+48);
while(*a.begin()=='0')
a.erase(a.begin());
if(a.size()==0)
a = "0";
return a;
}
string mul(string a,string b)
{
string c;
if(a.size()<b.size())
swap(a, b);
c = mul(a,b[b.size()-1]-'0');
string k = "0";
for(int i=b.size()-2;i>=0;i--)
{
string f = mul(a,b[i]-'0');
f = f+k;
c = add(c, f);
k = k + "0";
}
return c;
}