Ultraviolet A 10844-Bloques (second type of Stringer)

Ultraviolet A 10844-Bloques

Question Link

Question: given n numbers, the n numbers can be divided into subsets and divided into several methods.
Idea: At the beginning, I first thought about a state. dp [I] [j] indicates the solution for placing I numbers and dividing them into j sets. Then, from dp [I-1] [j-1] to one more set, and from dp [I-1] [j, the transfer equation is dp [I] [j] = dp [I-1] [j-1] + dp [I-1] [j] * j; it is reasonable to say that this state transfer is OK, but because the answer to this question is high precision, when n is 900, the answer is as high as more than 1700 bits. In addition, the high-precision calculation result times out, the bell number can be introduced through the Yang Hui triangle. Specifically, a number is equal to the sum of the Left and top left sides of the Yang Hui triangle, so that the State transfer becomes a pure high-precision addition, then the high-precision addition operation was performed to compress the bit with high precision, barely passing
Code:

#include 
  
   #include 
   
    #include 
    
     #include 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;}