A multiplication game
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6421 | Accepted: 3225 |
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.
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.
Source
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #define ll long long #define maxn 10005 using namespace std; ll n,tot,val; ll num[maxn],pos[20]; ll to[maxn][20]; bool win[maxn]; inline void prework(){ tot=0; for(ll i=1;i<n;i<<=1) for(ll j=i;j<n;j*=3ll) for(ll k=j;k<n;k*=5ll) for(ll l=k;l<n;l*=7ll) num[++tot]=l; sort(num+1,num+tot+1); memset(pos,0,sizeof(pos)); memset(win,0,sizeof(win)); for(int i=1;i<=tot;i++){ // to[i][1]=i; for(int j=2;j<=9;j++){ val=num[i]*(ll)j; while(pos[j]<tot&&num[pos[j]]<val) pos[j]++; to[i][j]=(num[pos[j]]==val?pos[j]:0); } } } int main(){ while(scanf("%lld",&n)==1&&n){ prework(); for(int i=tot;i;i--){ for(int j=2;j<=9;j++) if(!to[i][j]||(to[i][j]&&!win[to[i][j]])){ win[i]=1; break; } } if(win[1]) puts("Stan wins."); else puts("Ollie wins."); } return 0; }