[Original blog] poj 2505 a multiplication game combination game

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Question: there is a numberp=1, A and B take turns, and P can be multiplied by 2 ~ each time ~ A number in 9, givennAfter one person operatesp>=nThen the man wins and asks if the first hand wins.

• Winning status: there is a way to go to a losing status.
• Failed status: the successor status is a winning status.

We can know>=nThe numbers are all failed and can be transferred>=nThe smallest numbern/9(Rounded up), son/9~n-1All are competitive. Likewisen/9/2(All are rounded up) must be transferred to the smallestn/9~n-1(Must win) status, son/9/2~n/9-1To be defeated, so we can push it like this1, Take a look1Whether it is a winning or a defeated output.

PS.a/b(Rounded up) can be written(a-1)/b+1(Division ). Easy operation.

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring> //by zrt//problem:using namespace std;typedef long long LL;const int inf(0x3f3f3f3f);const double eps(1e-9); int main(){    #ifdef LOCAL    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);    #endif    LL n;    while(~scanf("%lld",&n)){        bool ok;        while(1){            LL p=(n-1)/9+1;            if(1>=p){                ok=1;break;            }else n=p;            p=(n-1)/2+1;            if(1>=p){                ok=0;break;            }else n=p;        }        if(ok){            puts("Stan wins.");        }else{            puts("Ollie wins. ");        }    }        return 0;}

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[Original blog] poj 2505 a multiplication game combination game