hdu 4873 ZCC Loves Intersection(大数+概率)

pid=4873" target="_blank" style="">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，每秒会依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

由于分子大于long long 的上限。所以在通分的时候为了避免大数相除，能够将分子差分成质因子来推断。

在高精度成低精度的时候要防止乘的过程中溢出int。由于N非常大。

#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;
}