rqnoj-116-质数取石子-dp

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
using namespace std;
#define maxn 20050
int 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]
!=-2)return dp[people]
;
    if(people==1)
    {
        int leap=0;
        dp[1]
=9999999;
        if(prime
||prime[n-1])
        {
            dp[1]
=1;
            return 1;
        }
        for(i=0;i<=nearprime
;i++)
        {
            if(pan(2,n-isprime[i])>-1)
            {
                dp[1]
=min(dp[1]
,dp[2][n-isprime[i]]+1);
                leap=1;
            }
        }
        if(leap==0)dp[1]
=-1;
    }
    else if(people==2)
    {
        dp[2]
=0;
        if(prime
||prime[n-1])
        {
            dp[2]
=-1;
            return -1;
        }
        for(i=0;i<=nearprime
;i++)
        {
            if(pan(1,n-isprime[i])==-1)
            {
                dp[2]
=-1;
                break;
            }
            else
            {
                dp[2]
=max(dp[1][n-isprime[i]]+1,dp[2]
);
            }
        }
    }
    return dp[people]
;
}
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;
}