hdu 1850 Being a Good Boy in Spring Festival(尼姆博弈加强)
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题意: 最后取完的选手胜利,与1907刚好相反。并输出若胜利,则第一步有多少种取的方式。
综上所述,若是 S2,S1,T0 。 则先下的人必输。
若是 T2,S0 。则先下的人必胜。
#include<iostream>#include<string.h>using namespace std;int main(){ int n , a[110] , m; int sum1 , sum2 , ans; while(scanf("%d",&n)!=EOF && n) { sum1=sum2=ans=m=0; for(int i=0;i<n;i++) { scanf("%d",&a[i]); if(a[i]>=2) sum1++; else sum2++; ans^=a[i]; } //cout<<"ans= "<<ans<<" sum1= "<<sum1<<" sum2= "<<sum2<<endl; int temp ; if(ans==0 && sum1>=2|| ans!=0&&sum1==0&&sum2%2!=0)//因为是最后取完者胜,所以判断定相反 printf("0\n"); if(ans!=0 && sum1!=0||ans==0 && sum1==0) { if(ans==0) { printf("1\n"); continue;} //cout<<"************"<<endl; for(int i=0;i<n;i++) { temp=ans; temp^=a[i]; //异或sum=0,且temp<a[i]说明可以通过取第i堆最后可取胜。 if(temp<a[i] && temp>0) m++; } printf("%d\n",m); } } return 0;}
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