HDU 6121 Build a tree [想法题]

题意：询问n个点的完全k叉树，所有子树节点个数的异或总和为多少。

题解：对于树的每一层，我们可以分为三种节点：①满节点的k叉树②不满的k叉树③比第一种情况少一层的满节点的k叉树，然后从叶子节点开始不断转移到上一层。

AC代码：

#include<vector>#include<list>#include<map>#include<set>#include<deque>#include<queue>#include<stack>#include<bitset>#include<algorithm>#include<functional>#include<numeric>#include<utility>#include<complex>#include<sstream>#include<iostream>#include<iomanip>#include<cstdio>#include<cmath>#include<cstdlib>#include<cstring>#include<ctime>#include<cassert>#include<string>using namespace std;typedef long long ll;int main(void){ll t;scanf("%lld",&t);while(t--){ll n,k;scanf("%lld%lld",&n,&k);if(k==1){ll t=n&3;if(t&1)printf("%lld\n",t/2u^1);else printf("%lld\n",t/2u^n); continue;}ll sum=1;ll a=0,b=0,c=0;ll s=k;ll flag=0;while(sum<=n-s){sum=sum+s;if((double)s>1e18/(double)k){flag=1;break;}s=s*k;}ll sheng=n-sum;if(flag==0)s/=k;a=sheng/k;b=sheng%k==0?0:1;c=s-a-b;ll ans=sheng%2;ll vala=k+1;ll valb;if(sheng%k!=0)valb=sheng%k+1;else valb=0;ll valc=1;while(a+b+c>1){if(a%2==1) ans=ans^vala;if(b%2==1) ans=ans^valb;if(c%2==1) ans=ans^valc;if(a%k!=0){valb=valb+vala*(a%k)+(k-b-a%k)*valc+1;b=1;}else valb=valb+(k-1)*valc+1;a=a/k; vala=vala*k+1;if(c%k!=0)b=1;c=(c-c%k)/k;valc=valc*k+1;}printf("%lld\n",ans^n);}return 0;}/*1 3  3      6 7 6 713131515131315152627262730*/