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;
}