UVA 10844 - Bloques (第二类斯特灵数)

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UVA 10844 - Bloques

题目链接

题意：给定n个数字，问这n个数字能分成子集分成有几种分法
思路：一开始先想了个状态，dp[i][j]表示放i个数字，分成j个集合的方案，那么转移为，从dp[i - 1][j - 1]在多一个集合，和从dp[i - 1][j]有j个位置放，那么转移方程为dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] * j;按理说这个状态转移是没问题的，但是由于这题答案是高精度，n为900时答案高达1700多位，加上高精度运算结果就超时了，后面知道这种是bell数，是可以通过杨辉三角推出来的，具体就是一个数字等于杨辉三角的左边和左上边相加，这样状态转移变成纯高精度加法，然后把高精度加法运算进行压缩位的高精度运算，勉强通过了
代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;using namespace std;
const int MAXN = 1800;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 int& b);bign operator + (const bign& b);bign operator - (const int& b);bign operator - (const bign& b);bign operator * (const int& b);bign operator * (const bign& b);bign operator / (const int& b);//bign operator / (const bign& b);int operator % (const int& b);
};/*Code*/const int N = 905;
int n;
bign dp[2][N], ans[N];void init() {int pre = 1, now = 0;dp[now][1] = 1; ans[1] = 1;for (int i = 2; i <= 900; i++) {swap(now, pre);dp[now][1] = dp[pre][i - 1];for (int j = 2; j <= i; j++)dp[now][j] = dp[now][j - 1] + dp[pre][j - 1];ans[i] = dp[now][i];}
}int main() {init();while (~scanf("%d", &n) && n) {printf("%d, ", n);ans[n].Put(); printf("\n");}return 0;
}void bign::DelZero () {while (len && num[len-1] == 0)len--;if (len == 0) {num[len++] = 0;}
}void bign::Put () {printf("%d", num[len - 1]);for (int i = len-2; i >= 0; i--) printf("%08d", num[i]);
}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 int& b) {bign a = b;return *this + a;
}bign bign::operator + (const bign& b) {int 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 % 100000000;bignSum /= 100000000;}while (bignSum) {ans.num[ans.len++] = bignSum % 100000000;bignSum /= 100000000;}return ans;
}bign bign::operator - (const int& b) {bign a = b;return *this - a;
}bign bign::operator - (const bign& b) {int 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;}ans.DelZero ();return ans;
}bign bign::operator * (const int& b) {int 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 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;}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;
}

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