hdu6121（想法题目）

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：

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1. #include<vector>  
2. #include<list>  
3. #include<map>  
4. #include<set>  
5. #include<deque>  
6. #include<queue>  
7. #include<stack>  
8. #include<bitset>  
9. #include<algorithm>  
10. #include<functional>  
11. #include<numeric>  
12. #include<utility>  
13. #include<complex>  
14. #include<sstream>  
15. #include<iostream>  
16. #include<iomanip>  
17. #include<cstdio>  
18. #include<cmath>  
19. #include<cstdlib>  
20. #include<cstring>  
21. #include<ctime>  
22. #include<cassert>  
23. #include<string>  
24. using namespace std;  
25. typedef long long ll;  
26. int main(void){  
27.     ll t;  
28.     scanf("%lld",&t);  
29.     while(t--){  
30.         ll n,k;  
31.         scanf("%lld%lld",&n,&k);  
32.         if(k==1)  
33.         {  
34.             ll t=n&3;  
35.             if(t&1)printf("%lld\n",t/2u^1);  
36.             else printf("%lld\n",t/2u^n);  
37.             continue;  
38.         }  
39.         ll sum=1;  
40.         ll a=0,b=0,c=0;  
41.         ll s=k;  
42.         ll flag=0;  
43.         while(sum<=n-s)  
44.         {  
45.             sum=sum+s;  
46.             if((double)s>1e18/(double)k)  
47.             {  
48.                 flag=1;  
49.                 break;  
50.             }  
51.             s=s*k;  
52.         }  
53.         ll sheng=n-sum;  
54.         if(flag==0)s/=k;  
55.         a=sheng/k;  
56.         b=sheng%k==0?0:1;  
57.         c=s-a-b;  
58.         ll ans=sheng%2;  
59.         ll vala=k+1;  
60.         ll valb;  
61.         if(sheng%k!=0)valb=sheng%k+1;  
62.         else valb=0;  
63.         ll valc=1;  
64.         while(a+b+c>1)  
65.         {  
66.             if(a%2==1) ans=ans^vala;  
67.             if(b%2==1) ans=ans^valb;  
68.             if(c%2==1) ans=ans^valc;  
69.             if(a%k!=0)  
70.             {  
71.                 valb=valb+vala*(a%k)+(k-b-a%k)*valc+1;  
72.                 b=1;  
73.             }  
74.             else valb=valb+(k-1)*valc+1;  
75.             a=a/k; vala=vala*k+1;  
76.             if(c%k!=0)b=1;  
77.             c=(c-c%k)/k;  
78.             valc=valc*k+1;  
79.               
80.         }  
81.         printf("%lld\n",ans^n);  
82.     }  
83.     return 0;  
84. }  
85.   
86. /*1 
87.  3  
88.  3      
89.  6 
90.  7 
91.  6 
92.  7 
93. 13 
94. 13 
95. 15 
96. 15 
97. 13 
98. 13 
99. 15 
100. 15 
101. 26 
102. 27 
103. 26 
104. 27 
105. 30 
106. */  

#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
 7
13
13
15
15
13
13
15
15
26
27
26
27
30
*/