题目大意:对于每一个位置来说,能够向上,水平,向下。坐标不能位负。每次上下移动最多为1。 给定n问说有多少种不同的图。结果对10100取模。
解题思路:由于最后都要落回y=0的位置,所以上升的次数和下降的次数是同样的,而且上升下降的关系满足出栈入栈的关系。即卡特兰数。
所以每次枚举i,表示有i个上升,i个下降,用组合数学枚举出位置,然后累加求和。
C(2∗in)∗f(i)=C(2∗i−2n)∗f(i−1)∗(n−2∗i+1)∗(n−2∗i+2)i∗(i+1)
注意取模后的前导0
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long type;
const int MAXN = 10005;
struct bign {
int len, num[MAXN];
bign () {
len = 0;
memset(num, 0, sizeof(num));
}
bign (int number) {*this = number;}
bign (const char* number) {*this = number;}
void DelZero ();
void Put ();
void operator = (int number);
void operator = (char* number);
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); }
void operator ++ ();
void operator -- ();
bign operator + (const type& b);
bign operator + (const bign& b);
bign operator - (const type& b);
bign operator - (const bign& b);
bign operator * (const type& b);
bign operator * (const bign& b);
bign operator / (const type& b);
//bign operator / (const bign& b);
int operator % (const int& b);
};
/*Code*/
int main () {
int n;
while (scanf("%d", &n) == 1) {
bign ans = 0;
bign tmp = 1;
ans = ans + tmp;
for (int i = 1; i <= n/2; i++) {
tmp = tmp * 1LL * (n - 2 * i + 2) * (n - 2 * i + 1);
tmp = tmp / (1LL * i * (i + 1));;
ans = ans + tmp;
ans.len = min(ans.len, 100);
}
ans.Put();
printf("
");
}
}
void bign::DelZero () {
while (len && num[len-1] == 0)
len--;
if (len == 0)
num[len++] = 0;
}
void bign::Put () {
bool flag = false;
for (int i = len-1; i >= 0; i--) {
if (num[i] || flag) {
printf("%d", num[i]);
flag = true;
}
}
}
void bign::operator = (char* number) {
len = strlen (number);
for (int i = 0; i < len; i++)
num[i] = number[len-i-1] - '0';
DelZero ();
}
void bign::operator = (int number) {
len = 0;
while (number) {
num[len++] = number%10;
number /= 10;
}
DelZero ();
}
bool bign::operator < (const bign& b) const {
if (len != b.len)
return len < b.len;
for (int i = len-1; i >= 0; i--)
if (num[i] != b.num[i])
return num[i] < b.num[i];
return false;
}
void bign::operator ++ () {
int s = 1;
for (int i = 0; i < len; i++) {
s = s + num[i];
num[i] = s % 10;
s /= 10;
if (!s) break;
}
while (s) {
num[len++] = s%10;
s /= 10;
}
}
void bign::operator -- () {
if (num[0] == 0 && len == 1) return;
int s = -1;
for (int i = 0; i < len; i++) {
s = s + num[i];
num[i] = (s + 10) % 10;
if (s >= 0) break;
}
DelZero ();
}
bign bign::operator + (const type& b) {
bign a = b;
return *this + a;
}
bign bign::operator + (const bign& b) {
type bignSum = 0;
bign ans;
for (int i = 0; i < len || i < b.len; i++) {
if (i < len) bignSum += num[i];
if (i < b.len) bignSum += b.num[i];
ans.num[ans.len++] = bignSum % 10;
bignSum /= 10;
}
while (bignSum) {
ans.num[ans.len++] = bignSum % 10;
bignSum /= 10;
}
return ans;
}
bign bign::operator - (const type& b) {
bign a = b;
return *this - a;
}
bign bign::operator - (const bign& b) {
type bignSub = 0;
bign ans;
for (int i = 0; i < len || i < b.len; i++) {
bignSub += num[i];
bignSub -= b.num[i];
ans.num[ans.len++] = (bignSub + 10) % 10;
if (bignSub < 0) bignSub = -1;
else bignSub = 0;
}
ans.DelZero ();
return ans;
}
bign bign::operator * (const type& b) {
type bignSum = 0;
bign ans;
ans.len = len;
for (int i = 0; i < len; i++) {
bignSum += num[i] * b;
ans.num[i] = bignSum % 10;
bignSum /= 10;
}
while (bignSum) {
ans.num[ans.len++] = bignSum % 10;
bignSum /= 10;
}
return ans;
}
bign bign::operator * (const bign& b) {
bign ans;
ans.len = 0;
for (int i = 0; i < len; i++){
int bignSum = 0;
for (int j = 0; j < b.len; j++){
bignSum += num[i] * b.num[j] + ans.num[i+j];
ans.num[i+j] = bignSum % 10;
bignSum /= 10;
}
ans.len = i + b.len;
while (bignSum){
ans.num[ans.len++] = bignSum % 10;
bignSum /= 10;
}
}
return ans;
}
bign bign::operator / (const type& b) {
bign ans;
type s = 0;
for (int i = len-1; i >= 0; i--) {
s = s * 10 + num[i];
ans.num[i] = s/b;
s %= b;
}
ans.len = len;
ans.DelZero ();
return ans;
}
int bign::operator % (const int& b) {
bign ans;
int s = 0;
for (int i = len-1; i >= 0; i--) {
s = s * 10 + num[i];
ans.num[i] = s/b;
s %= b;
}
return s;
}