hdu 4873 ZCC Loves Intersection(大数+概率)

题目链接：hdu 4873 ZCC Loves Intersection

题目大意：给出N，D，表示在一个D维的坐标上，坐标范围为0~N-1。在这个坐标系中有D条线段，分别平行与各个坐标轴，每秒会根据题目中的伪代码随机生成各个线段的位置。两条直线相交的话会得一分，问每秒得分的期望。

解题思路：总的情况(ND−1∗C(2N))D,两条直线相交的得分C(2D)∗s2∗ND−2∗(ND−2∗C(2N))D−2
s是在二维情况下的分的情况数s=∑i=1N((N−i+1) i−1)=N3+3∗N2−4∗N6
最后化简成ans=C(2D)∗(N+4)29∗ND
因为分子大于long long 的上限，所以在通分的时候为了避免大数相除，可以将分子差分成质因子来判断。
在高精度成低精度的时候要防止乘的过程中溢出int，因为N很大。

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>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;}