題意：給你n個木塊，讓你搭建筑物，可以幾個木塊一起搭，也可以一個成建筑物，問有多少種方法？

題解：就是斯特靈數模型 ?求和；（即貝爾數）

/******************************************************* * 貝爾數模型 * * 1 * 1 2 * 2 3 5 * 5 7 10 15 * 15 20 27 37 52 * . . . . . * . . . . . * 第1列即是貝爾數，看圖找規律，祝你成功 **********************************************************/ #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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