A Multiplication Game

Problem Description

Stan and Ollie play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Stan always starts with p = 1, does his multiplication, then Ollie multiplies the number, then Stan and so on. Before a game starts, they draw an integer
1 < n < 4294967295 and the winner is who first reaches p >= n.

Input

Each line of input contains one integer number n.

Output

For each line of input output one line either

Stan wins.

or

Ollie wins.

assuming that both of them play perfectly.

Sample Input

162
17
34012226

Sample Output

Stan wins.
Ollie wins.
Stan wins.

题目大意：

Stan 和 Ollie 两个人玩游戏， 一开始数字是 1，两个人轮流，Stan 先手，每个人选择 2~9 里面的一个一直乘以起初的数字，直到某个人乘到 这个数大于等于 n 就算赢，n给定，问你谁必然赢。

解题思路：

这题我没找出SG函数（必胜必输）的规律，只能用DP的方法求出每种状态的必胜必输状态。

2~9 ，依次其实可以用 2 3  5  7 这几个数得到，因此每个状态只需要记录2 3  5  7 这几个数的数目即可，接下来就是2~9的偏移量，可以 这么表示

const int off2[]={1,0,2,0,1,0,3,0};

const int off3[]={0,1,0,0,1,0,0,2};

const int off5[]={0,0,0,1,0,0,0,0};

const int off7[]={0,0,0,0,0,1,0,0};

解题代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

typedef long long ll;

const int off2[]={1,0,2,0,1,0,3,0};
const int off3[]={0,1,0,0,1,0,0,2};
const int off5[]={0,0,0,1,0,0,0,0};
const int off7[]={0,0,0,0,0,1,0,0};

int dp[40][40][40][40][2];
ll n;

ll pow_mod(ll a,int b){
    ll ans=1,tmp=a;
    while( b>0 ){
        if( b%2!=0 ) ans*=tmp;
        b/=2;
        tmp*=tmp;
    }
    return ans;
}

int DP(int c2,int c3,int c5,int c7,int f){
    ll tmp=pow_mod(2,c2)*pow_mod(3,c3)*pow_mod(5,c5)*pow_mod(7,c7);
    if(tmp>=n) return 1-f;
    if(dp[c2][c3][c5][c7][f]!=-1) return dp[c2][c3][c5][c7][f];
    int ans;
    if(f==0){
        ans=1;
        for(int i=0;i<8;i++){
            int cc2=c2+off2[i],cc3=c3+off3[i],cc5=c5+off5[i],cc7=c7+off7[i];
            if(DP(cc2,cc3,cc5,cc7,1-f)<ans){
                ans=0;
                break;
            }
        }
    }else{
        ans=0;
        for(int i=0;i<8;i++){
            int cc2=c2+off2[i],cc3=c3+off3[i],cc5=c5+off5[i],cc7=c7+off7[i];
            if(DP(cc2,cc3,cc5,cc7,1-f)>ans){
                ans=1;
                break;
            }
        }
    }
    return dp[c2][c3][c5][c7][f]=ans;
}

int main(){
    while(cin>>n){
        memset(dp,-1,sizeof(dp));
        if(DP(0,0,0,0,0)==0) printf("Stan wins.\n");
        else printf("Ollie wins.\n");
    }
    return 0;
}

HDU 1517 A Multiplication Game （博弈-求sg）