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;}