rqnoj-116-质数取石子-dp

dp[1][i]:

DD选择的时候，剩余i石子，DD赢的最小步数。若此时DD是必败态，则为-1；

dp[2][i]:

MM选择的时候，剩余i石子，MM输的最大步数。若此时DD是必胜态，则为-1；

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<math.h>using namespace std;#define maxn 20050int prime[maxn];int isprime[maxn];int nearprime[maxn];int numprime;int dp[3][maxn];void dos(){    prime[0]=0;    prime[1]=0;    prime[2]=1;    for(int i=3;i<maxn;i++)prime[i]=i%2;    for(int i=2;i<sqrt(maxn);i++)    {        if(prime[i]==0)continue;        for(int j=2;j*i<maxn;j++)        {            prime[j*i]=0;        }    }    numprime=0;    for(int i=0;i<maxn;i++)    {        if(prime[i])isprime[numprime++]=i;        nearprime[i]=numprime-1;    }}int pan(int people,int n){    int i;    if(dp[people][n]!=-2)return dp[people][n];    if(people==1)    {        int leap=0;        dp[1][n]=9999999;        if(prime[n]||prime[n-1])        {            dp[1][n]=1;            return 1;        }        for(i=0;i<=nearprime[n];i++)        {            if(pan(2,n-isprime[i])>-1)            {                dp[1][n]=min(dp[1][n],dp[2][n-isprime[i]]+1);                leap=1;            }        }        if(leap==0)dp[1][n]=-1;    }    else if(people==2)    {        dp[2][n]=0;        if(prime[n]||prime[n-1])        {            dp[2][n]=-1;            return -1;        }        for(i=0;i<=nearprime[n];i++)        {            if(pan(1,n-isprime[i])==-1)            {                dp[2][n]=-1;                break;            }            else            {                dp[2][n]=max(dp[1][n-isprime[i]]+1,dp[2][n]);            }        }    }    return dp[people][n];}void play(){    dos();    int i;    for(i=0;i<maxn;i++)dp[1][i]=dp[2][i]=-2;    dp[1][0]=-1;    dp[1][1]=-1;    dp[1][2]=1;    dp[2][0]=0;    dp[2][1]=0;    dp[2][2]=-1;    for(i=3;i<maxn;i++)    {        dp[1][i]=pan(1,i);    }}int main(){    int n,m;    play();    scanf("%d",&n);    while(n--)    {        scanf("%d",&m);        printf("%d\n",dp[1][m]);    }    return 0;}