ZOJ

题目：2-20这19个数字的游戏。每取走一个数之后，这个数的倍数便不能再取，而且某两个取过的数的倍数的和，也不能再取。给出当前可取数字的状态，问当前状态是不是先手必胜状态。如果是，输出取法

思路：利用NP状态定理和状态压缩做。

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3f//0x3f3f3f3fconst int maxn=(1<<19)+5;int sg[maxn];int get_state(int state,int x){//取x这个数    int res=state;    for(int i=x;i<=20;i+=x)//把x的倍数去掉        res&=~(1<<(i-2));    for(int i=2;i<=20;i++){        if((res&(1<<(i-2)))){            for(int j=x;i-j-2>=0;j+=x){//把k*x和某个数的和去掉                if(!(res&(1<<(i-j-2)))){                    res&=~(1<<(i-2));                    break;                }            }        }    }    return res;}int get_sg(int state){    if(sg[state]!=-1)        return sg[state];    sg[state]=0;    for(int i=2;i<=20;i++){        if(state&(1<<(i-2))){            int tmp=get_state(state,i);            if(get_sg(tmp)==0){                sg[state]=1;                break;            }        }    }    return sg[state];}int a[25];int main(){//    freopen("D:\\input.txt","r",stdin);//    freopen("D:\\output.txt","w",stdout);    int T,cas=0,n;    mm(sg,-1);    scanf("%d",&T);    while(T--){        scanf("%d",&n);        int state=0;        for(int i=0;i<n;i++){            scanf("%d",&a[i]);            state|=(1<<(a[i]-2));        }        printf("Scenario #%d:\n",++cas);        if(get_sg(state)==0)            printf("There is no winning move.\n");        else{            printf("The winning moves are:");            for(int i=0;i<n;i++){                int tmp=get_state(state,a[i]);                if(get_sg(tmp)==0)                    printf(" %d",a[i]);            }            printf(".\n");        }        printf("\n");    }    return 0;}