Problem B
Complete Tree Labeling!
Input: standard input
Output: standard output
Time Limit: 45 seconds
Memory Limit: 32 MB
A complete k-ary tree is a k-ary tree in which all leaves have same depth and all internal nodes have degree k. This k is also known as the branching factor of a tree. It is very easy to determine the number of nodes of such a tree. Given the depth and branching factor of such a tree, you will have to determine in how many different ways you can number the nodes of the tree so that the label of each node is less that that of its descendants. You should assume that for numbering a tree with N nodes you have the (1, 2, 3, N-1, N) labels available.
Input
The input file will contain several lines of input. Each line will contain two integers k and d. Here k is the branching factor of the complete k-arytree and d is the depth of the complete k-ary tree (k>0, d>0, k*d<=21).
Output
For each line of input, produce one line of output containing a round number, which is the number of ways the k-ary tree can be labeled, maintaining the constraints described above.
Sample Input:
2 2
10 1
Sample Output:
80
3628800
题意：k叉d层树最多组成几种搜索树。
思路：参考http://www.2cto.com/kf/201310/251470.html
代码：
#include <stdio.h>
#include <string.h>
#include <math.h>
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
const int MAXSIZE = 10000;
struct bign {
int s[MAXSIZE];
bign () {memset(s, 0, sizeof(s));}
bign (int number) {*this = number;}
bign (const char* number) {*this = number;}
void put();
bign mul(int d);
void del();
void init() { memset(s, 0, sizeof(s)); }
bign operator = (char *num);
bign operator = (int num);
bool operator < (const bign& b) const;
bool operator > (const bign& b) const { return b < *this; }
bool operator <= (const bign& b) const { return !(b < *this); }
bool operator >= (const bign& b) const { return !(*this < b); }
bool operator != (const bign& b) const { return b < *this || *this < b;}
bool operator == (const bign& b) const { return !(b != *this); }
bign operator + (const bign& c);
bign operator * (const bign& c);
bign operator - (const bign& c);
int operator / (const bign& c);
bign operator / (int k);
bign operator % (const bign &c);
int operator % (int k);
void operator ++ ();
bool operator -- ();
};
bign f[25][25];
int node[25][25];
int n, m;
bign c(int n, int m) {
bign ans = 1;
m = min(m, n - m);
for (int i = 0; i < m; i ++) {
bign save = (n - i);
ans = ans * save / (i + 1);
}
return ans;
}
void init() {
int i, j, k;
for (i = 1; i <= 21; i ++) {
f[i][0] = 1; node[i][0] = 1;
for (j = 1; j <= 21 / i; j ++) {
f[i][j] = 1;
for (k = 0; k < i; k ++) {
node[i][j] = node[i][j - 1] * i + 1;
f[i][j] = f[i][j] * c(node[i][j] - 1 - k * node[i][j - 1], node[i][j - 1]) * f[i][j - 1];
}
}
}
}
int main() {
init();
while (~scanf("%d%d", &n, &m)) {
f[n][m].put();
printf("
");
}
return 0;
}
bign bign::operator = (char *num) {
init();
s[0] = strlen(num);
for (int i = 1; i <= s[0]; i++)
s[i] = num[s[0] - i] - '0';
return *this;
}
bign bign::operator = (int num) {
char str[MAXSIZE];
sprintf(str, "%d", num);
return *this = str;
}
bool bign::operator < (const bign& b) const {
if (s[0] != b.s[0])
return s[0] < b.s[0];
for (int i = s[0]; i; i--)
if (s[i] != b.s[i])
return s[i] < b.s[i];
return false;
}
bign bign::operator + (const bign& c) {
int sum = 0;
bign ans;
ans.s[0] = max(s[0], c.s[0]);
for (int i = 1; i <= ans.s[0]; i++) {
if (i <= s[0]) sum += s[i];
if (i <= c.s[0]) sum += c.s[i];
ans.s[i] = sum % 10;
sum /= 10;
}
return ans;
}
bign bign::operator * (const bign& c) {
bign ans;
ans.s[0] = 0;
for (int i = 1; i <= c.s[0]; i++){
int g = 0;
for (int j = 1; j <= s[0]; j++){
int x = s[j] * c.s[i] + g + ans.s[i + j - 1];
ans.s[i + j - 1] = x % 10;
g = x / 10;
}
int t = i + s[0] - 1;
while (g){
++t;
g += ans.s[t];
ans.s[t] = g % 10;
g = g / 10;
}
ans.s[0] = max(ans.s[0], t);
}
ans.del();
return ans;
}
bign bign::operator - (const bign& c) {
bign ans = *this;
int i;
for (i = 1; i <= c.s[0]; i++) {
if (ans.s[i] < c.s[i]) {
ans.s[i] += 10;
ans.s[i + 1]--;;
}
ans.s[i] -= c.s[i];
}
for (i = 1; i <= ans.s[0]; i++) {
if (ans.s[i] < 0) {
ans.s[i] += 10;
ans.s[i + 1]--;
}
}
ans.del();
return ans;
}
int bign::operator / (const bign& c) {
int ans = 0;
bign d = *this;
while (d >= c) {
d = d - c;
ans++;
}
return ans;
}
bign bign::operator / (int k) {
bign ans;
ans.s[0] = s[0];
int num = 0;
for (int i = s[0]; i; i--) {
num = num * 10 + s[i];
ans.s[i] = num / k;
num = num % k;
}
ans.del();
return ans;
}
int bign:: operator % (int k){
int sum = 0;
for (int i = s[0]; i; i--){
sum = sum * 10 + s[i];
sum = sum % k;
}
return sum;
}
bign bign::operator % (const bign &c) {
bign now = *this;
while (now >= c) {
now = now - c;
now.del();
}
return now;
}
void bign::operator ++ () {
s[1]++;
for (int i = 1; s[i] == 10; i++) {
s[i] = 0;
s[i + 1]++;
s[0] = max(s[0], i + 1);
}
}
bool bign::operator -- () {
del();
if (s[0] == 1 && s[1] == 0) return false;
int i;
for (i = 1; s[i] == 0; i++)
s[i] = 9;
s[i]--;
del();
return true;
}
void bign::put() {
if (s[0] == 0)
printf("0");
else
for (int i = s[0]; i; i--)
printf("%d", s[i]);
}
bign bign::mul(int d) {
s[0] += d;
int i;
for (i = s[0]; i > d; i--)
s[i] = s[i - d];
for (i = d; i; i--)
s[i] = 0;
return *this;
}
void bign::del() {
while (s[s[0]] == 0) {
s[0]--;
if (s[0] == 0) break;
}
}