HDU 4873 ZCC Loves Intersection (probability), hdu4873

HDU 4873 ZCC Loves Intersection (probability), hdu4873
HDU 4873 ZCC Loves Intersection

Question Link

In d-dimension blocks with a length of n, d lines parallel to each axis are selected each time. If there is a pair of intersection, the number of points is added to 1, ask the expected number of points each time

Train of Thought: I think it is still a little worse. I have to turn to the official question and answer it. I feel that I did not expect to split the numerator into the sum of several polynomials, and then I can convert it into a formula for solving the problem.

Code:

#include <cstdio>#include <cstring>#include <cmath>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);};/***************************************************/const int N = 10005;long long n, d, prime[N], cnt[N];int pn = 0, vis[N];bign zi, mu;void table() {    for (long long i = 2; i < N; i++) {prime[pn++] = i;for (long long j = i * i; j < N; j += i)    vis[j] = 1;    }}bign qpow(long long x, long long k) {    bign ans = 1;    bign tmp = x;    while (k) {if (k&1) ans = ans * tmp;tmp = tmp * tmp;k >>= 1;    }    return ans;}void solve(long long num, long long val) {    for (int i = 0; i < pn && prime[i] <= num; i++) {while (num % prime[i] == 0) {    cnt[i] += val;    num /= prime[i];}    }    if (num != 1) {if (val > 0)    zi = zi * qpow(num, val);else if (val < 0)    mu = mu * qpow(num, (-val));    }}int main() {    table();    while (~scanf("%lld%lld", &n, &d)) {zi = 1, mu = 1;memset(cnt, 0, sizeof(cnt));solve(d * (d - 1) / 2, 1);solve(n + 4, 2);solve(3, -2);solve(n, -d);for (int i = 0; i < pn; i++) {    if (cnt[i] > 0)zi = zi * qpow(prime[i], cnt[i]);    else if (cnt[i] < 0)mu = mu * qpow(prime[i], (-cnt[i]));}zi.Put();if (mu != 1) {    printf("/");    mu.Put();}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;}