URAL 2068 Game of Nuts(博弈)
2016-08-22 19:36
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题目地址:http://acm.timus.ru/problem.aspx?space=1&num=2068
思路:x个石子,x=2*k+1。分三堆2*k+1=(2*k1+1)+(2*k2+1)+(2*k3+1)=2(k1+k2+k3+1)+1。k1+k2+k3+1=k,每分一次,k1+k2+k3减少1,所以石子最多分k次。求出总次数,判断奇偶即可。
思路:x个石子,x=2*k+1。分三堆2*k+1=(2*k1+1)+(2*k2+1)+(2*k3+1)=2(k1+k2+k3+1)+1。k1+k2+k3+1=k,每分一次,k1+k2+k3减少1,所以石子最多分k次。求出总次数,判断奇偶即可。
#include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define debu using namespace std; int main() { #ifdef debug freopen("in.in","r",stdin); #endif // debug int n; while(scanf("%d",&n)==1) { int sum=0; for(int i=0;i<n;i++) { int x; scanf("%d",&x); sum+=x/2; } if(!(sum&1)) printf("Stannis\n"); else printf("Daenerys\n"); } return 0; }
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