HDU 1848 Fibonacci again and again(博弈 SG运用)
2017-08-09 10:20
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有3堆石头,每堆石头可以操作的方法都一样,且相互之间不会影响。
根据SG定理,把他分为3堆独立的石头。求SG函数,最后异或起来
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
int f[20],s[1005],sg[1005];
void GetSG(int n)
{
int i,j;
memset(sg,0,sizeof(sg));
for(i=1;i<=n;i++)
{
memset(s,0,sizeof(s));
for(j=1;f[j]<=i&&j<=16;j++)
s[sg[i-f[j]]]=1;
for(j=0; ;j++)
if(!s[j])
{
sg[i]=j;
break;
}
}
}
int main()
{
int i,n,m,k;
f[0]=1;f[1]=1;f[2]=2;
for(i=3;i<=16;i++)//17已经超出范围
f[i]=f[i-1]+f[i-2];
GetSG(1000);
while(scanf("%d%d%d",&n,&m,&k),n&&m&&k)
{
if(sg
^sg[m]^sg[k])
printf("Fibo\n");
else
printf("Nacci\n");
}
}
根据SG定理,把他分为3堆独立的石头。求SG函数,最后异或起来
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
int f[20],s[1005],sg[1005];
void GetSG(int n)
{
int i,j;
memset(sg,0,sizeof(sg));
for(i=1;i<=n;i++)
{
memset(s,0,sizeof(s));
for(j=1;f[j]<=i&&j<=16;j++)
s[sg[i-f[j]]]=1;
for(j=0; ;j++)
if(!s[j])
{
sg[i]=j;
break;
}
}
}
int main()
{
int i,n,m,k;
f[0]=1;f[1]=1;f[2]=2;
for(i=3;i<=16;i++)//17已经超出范围
f[i]=f[i-1]+f[i-2];
GetSG(1000);
while(scanf("%d%d%d",&n,&m,&k),n&&m&&k)
{
if(sg
^sg[m]^sg[k])
printf("Fibo\n");
else
printf("Nacci\n");
}
}
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