uva 10516 - Another Counting Problem(dp)

题目链接：uva 10516 - Another Counting Problem

题目大意：给出n和d，计算出有多少n叉树层数为d层，对于每个节点，要不为空，要不为满。

解题思路：c[i]表示说层数不大于d的n叉数的个数，c[i] = c[i - 1] ^ n + 1, 最后答案ans = c[d] - c[d - 1].

#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 = 1005;

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();

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 c[18];

int main () {

int n, d;
while (scanf("%d%d", &n, &d) == 2 && n) {
printf("%d %d ", n, d);

if (d == 0) printf("1\n");
else {

c[0] = 1;
for (int i = 1; i <= d; i++) {
c[i] = 1;
for (int j = 0; j < n; j++)
c[i] = c[i] * c[i - 1];
++c[i];
}
bign ans = c[d] - c[d - 1];
ans.put();
printf("\n");
}
}
return 0;
}

bign bign::operator = (char *num) {
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;
for (int 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 (int 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;
for (int i = s[0]; i > d; i--)
s[i] = s[i - d];
for (int i = d; i; i--)
s[i] = 0;
return *this;
}

void bign::del() {
while (s[s[0]] == 0) {
s[0]--;
if (s[0] == 0) break;
}
}