HDU 4873 ZCC Loves Intersection

题目链接

题意：d维的。长度为n的块中，每次选d条平行于各条轴的线段，假设有两两相交则点数加1，问每次得到点数的期望是多少

思路：自己推还是差一些，转篇官方题接把，感觉自己想的没想到把分子那项拆分成几个多项式的和，然后能够转化为公式求解。

代码：

#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;
}