2017 Multi-University Training Contest

其他题目题解：

2017 Multi-University Training Contest - Team 7：1005. Euler theorem

2017 Multi-University Training Contest - Team 7：1011. Kolakoski

2017 Multi-University Training Contest - Team 7：1008. Hard challenge

2017 Multi-University Training Contest - Team 7：1003. Color the chessboard

2017 Multi-University Training Contest - Team 7：1010. Just do it

题意：有一棵n个点的有根数，根编号为0，i号节点的父亲是(i-1)/k号节点

求出所有子树大小的异或和

性质：对于一颗满k叉树而言，如果k是偶数，那么它的异或和就是树的大小

如果k是奇数，那么它的异或和就是任意一棵子树大小的异或和再异或树的大小

而这题的树也有几个性质：

①假设n足够大，那么根有k个儿子（废话）

②这k棵子树最多只有一棵不是满二叉树

那么假设k不等于1，二叉树的深度不会超过63

所以只要找到是哪棵子树不是满二叉树，以那棵子树作为整棵树求它的异或和，很显然子树也具有一样的性质，这样就可以递归求解，求出来之后直接异或其它所有满二叉树的大小就是答案，注意别忘记异或上n

还有k=1要特判

#include<stdio.h>#define LL long longLL tre[125] = {1};int main(void){LL T, n, k, ans, now, last, x, l, r, full;scanf("%lld", &T);while(T--){scanf("%lld%lld", &n, &k);if(k==1){switch(n%4){case 1:  printf("1\n");  break;case 2:  printf("%lld\n", n+1);  break;case 3:  printf("0\n");  break;case 0:  printf("%lld\n", n);  break;}}else{ans = n;while(1){if(n-1<=k){if(n%2==0)ans ^= 1;printf("%lld\n", ans);break;}now = x = full = 1;while(0<=now+x*k && now+x*k<n){x = x*k;now += x;full ^= now;}last = n-now;l = (last-1)/x;r = k-l-1;if(k%2==0){if(l%2==1)  ans ^= now;if(r%2==1)  ans ^= (now-x);}else{if(l%2==1)  ans ^= full;if(r%2==1)  ans ^= full^now;}n -= l*now+r*(now-x)+1;ans ^= n;}}}return 0;}