Hdu 4873 ZCC Loves Intersection (large number + probability)

Hdu 4873 ZCC Loves Intersection (large number + probability)

Link: hdu 4873 ZCC Loves Intersection

Given N, D, indicates that the coordinate range of a D is 0 ~ N-1. In this coordinate system, there are D line segments parallel to each coordinate axis, and the positions of each line segment are randomly generated per second based on the pseudocode in the question. If two straight lines intersect, you will get a point and ask the expected score per second.

Solution: General situation (ND? 1? C (2N) D: the intersection of two straight lines. C (2D )? S2? ND? 2? (ND? 2? C (2N) D? 2
S is the number of points in two-dimensional conditions. s = Σ I = 1N (N? I + 1) I? 1) = N3 + 3? N2? 4? N6
Is it simplified to ans = C (2D )? (N + 4) 29? ND
Because molecules are larger than the upper limit of long, in order to avoid division of large numbers, the molecular difference can be determined by dividing them into quality factors.
When the precision is low, int overflow should be prevented in the multiplication process, because N is very large.

#include 
  
   #include 
   
    #include 
    
     #include 
     
      using namespace std;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 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 maxn = 5000;int cnt, num[maxn];void divfact (int n) {    int m = (int)sqrt(n + 0.5);    for (int i = 2; i <= m; i++) {        if (n % i)            continue;        num[cnt++] = i;        while (n % i == 0)            n /= i;    }    if (n != 1)        num[cnt++] = n;}bign power (bign x, int d) {    bign ans = 1;    while (d) {        if (d & 1)            ans = ans * x;        x = x * x;        d /= 2;    }    return ans;}int main () {    int N, D;    while (scanf("%d%d", &N, &D) == 2 && N + D) {        cnt = 0;        bign q = N;        q = power(q, D) * 9;        bign p = D * (D - 1) / 2;        p = p * (N + 4);        p = p * (N + 4);        divfact(N);        num[cnt++] = 3;        for (int i = 0; i < cnt; i++) {            while (p % num[i] == 0 && q % num[i] == 0) {                p = p / num[i];                q = q / num[i];            }        }        if (p != q) {            p.Put();            printf("/");            q.Put();        } else            printf("1");        printf("\n");    }    return 0;}void bign::DelZero () {    while (len && num[len-1] == 0)        len--;    if (len == 0) {        num[len++] = 0;    }}void bign::Put () {    for (int i = len-1; i >= 0; i--)         printf("%d", 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 % 10;        bignSum /= 10;    }    while (bignSum) {        ans.num[ans.len++] = bignSum % 10;        bignSum /= 10;    }    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) {    long long bignSum = 0;    bign ans;    ans.len = len;    for (int i = 0; i < len; i++) {        bignSum += (long long)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;}