Good Luck in CET-4 Everybody! HDU - 1847
题意:一堆n个的牌,每次抽1,2,4,8(2^k)张,最后不能取的输
思路:
直接手写博弈树,发现3的倍数一定为必败态.大胆猜想必败态,借着接着从3个性质证明
1. 当n=0(终止状态) 必败
2.对于当前局面是必败的,任何操作都只能转移到必胜态
3.对于当前局面必胜,可以转移到必败态
证明:
1.显然成立
2. 即
因为 2^k不可能有因数3,因此一定不能整除,所以对于当前为必败态的局面,任何操作都只能转为必胜态
3. 对于一个数n , 我们要证明,在n%3!=0的情况下, 一定存在一个k, 使得 . 打表表明在n <= 1000 的情况下一定成立.
等价
n%3=0,1,2 2^k%3=0,1,2 证毕
看了下网上说的是Bash博弈,想想有道理. 但是大多数博客对结论没证明
对于递归和非递归的版本, sg上限是 1000
#include<cstdio>
#include<vector>
#include<cmath>
#include<math.h>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<map>
#define PI acos(-1.0)
#define pb push_back
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=1005;
const int MOD=1e9+7;
const double EPS=1e-10;
template <class T>
bool sf(T &ret){ //Faster Input
char c; int sgn; T bit=0.1;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
if(c==' '||c=='\n'){ ret*=sgn; return 1; }
while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
ret*=sgn;
return 1;
}
int main(void){
int n;
while(scanf("%d",&n)==1){
if(n%3==0) printf("Cici\n");
else printf("Kiki\n");
}
}作为第一题博弈题,再写一遍get_sg和dfs版本
//SG函数版本
#include<cstdio>
#include<vector>
#include<cmath>
#include<math.h>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<map>
#define PI acos(-1.0)
#define pb push_back
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=10005;
const int MOD=1e9+7;
const double EPS=1e-10;
template <class T>
bool sf(T &ret){ //Faster Input
char c; int sgn; T bit=0.1;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
if(c==' '||c=='\n'){ ret*=sgn; return 1; }
while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
ret*=sgn;
return 1;
}
int sg[N];
bool vis[N];
int f[N];
void get_sg(){
int t=1;
for(int i=1;i<=20;i++){
f[i]=t;
t*=2;
}
memset(sg,0,sizeof sg);
sg[0]=0;
for(int n=1;n<=1000;n++){
memset(vis,0,sizeof vis);
for(int j=1;j<=20&&n>=f[j];j++) vis[sg[n-f[j]]]=1;
for(int i=0;i<=1000;i++){
if(!vis[i]){
sg[n]=i;
break;
}
}
}
return ;
}
int main(void){
get_sg();
int n;
while(scanf("%d",&n)==1){
if(sg[n]) printf("Kiki\n");
else printf("Cici\n");
}
return 0;
}
//sg_dfs版
#include<cstdio>
#include<vector>
#include<cmath>
#include<math.h>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<map>
#define PI acos(-1.0)
#define pb push_back
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=10005;
const int MOD=1e9+7;
const double EPS=1e-10;
template <class T>
bool sf(T &ret){ //Faster Input
char c; int sgn; T bit=0.1;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
if(c==' '||c=='\n'){ ret*=sgn; return 1; }
while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
ret*=sgn;
return 1;
}
int sg[N];
int f[N];
int sg_dfs(int n){
if(sg[n]!=-1) return sg[n];
bool vis[50];
memset(vis,false,sizeof vis);
for(int i=1;i<=30&&f[i]<=n;i++){
sg_dfs(n-f[i]);
vis[sg[n-f[i]]]=1;
}
for(int i=0;i<=1000;++i){
if(!vis[i]){
return sg[n]=i;
}
}
}
int main(void){
memset(sg,-1,sizeof sg);
int t=1;
for(int i=1;i<=20;i++) f[i]=t,t*=2;
sg_dfs(1000);
int n;
while(scanf("%d",&n)==1){
if(sg[n]) printf("Kiki\n");
else printf("Cici\n");
}
return 0;
}