hdu 4762 Cut the Cake（推导+高精度）

题目链接：hdu 4762 Cut the Cake

题目大意：给出m和n，表示说在一块蛋糕上要分成m份，然后有n个草莓，随机分布在蛋糕上，现在你为切蛋糕的人，你想要尽量多的草莓，问说你获得所有草莓的概率。

解题思路：p = n*（1/m)^(n-1)，要用到高精度，公式推导：首先假设i号草莓为切点，那么有n种可能（每个草莓都有可能作为切点，即极角最小的），那么该种情况下，想要将所有草莓占为己有的可能即为剩下n-1个草莓全分布在该草莓极角大的一侧的1/m的范围内。因为m，n>1，所以肯定至少切一刀，注意答案要化简。

#include <stdio.h>
#include <string.h>
#include <iostream>

using namespace std;
const int N = 1005;

struct bign {
	int len, sex;
	int s[N];

	bign() {
		this -> len = 1;
		this -> sex = 0;
		memset(s, 0, sizeof(s));
	}


	bign operator = (const char *number) {
		int begin = 0;
		len = 0;
		sex = 1;
		if (number[begin] == '-') {
			sex = -1;
			begin++;
		}
		else if (number[begin] == '+')
			begin++;

		for (int j = begin; number[j]; j++)
			s[len++] = number[j] - '0';
		return *this;
	}
	bign operator = (int number) {
		char string[N];
		sprintf(string, "%d", number);
		*this = string;
		return *this;
	}
	bign (int number) {*this = number;}
	bign (const char* number) {*this = number;}

	bign change(bign cur) {
		bign now;
		now = cur;
		for (int i = 0; i < cur.len; i++)
			now.s[i] = cur.s[cur.len - i - 1];
		return now;
	}

	void delZore() {	// 删除前导0.
		bign now = change(*this);
		while (now.s[now.len - 1] == 0 && now.len > 1) {
			now.len--;
		}
		*this = change(now);
	}

	void put() {    // 输出数值。
		delZore();
		if (sex < 0 && (len != 1 || s[0] != 0))
			cout << "-";
		for (int i = 0; i < len; i++)
			cout << s[i];
	}

	bign operator * (const bign &cur){  
		bign sum, a, b;
		sum.len = 0; 
		a = a.change(*this);
		b = b.change(cur);

		for (int i = 0; i < a.len; i++){  
			int g = 0;  

			for (int j = 0; j < b.len; j++){  
				int x = a.s[i] * b.s[j] + g + sum.s[i + j];  
				sum.s[i + j] = x % 10;  
				g = x / 10;  
			}  
			sum.len = i + b.len;  

			while (g){  
				sum.s[sum.len++] = g % 10;  
				g = g / 10;  
			}  
		}  
		return sum.change(sum);  
	}  

	bign operator / (int k) {  // 高精度求商低精度。
		bign sum;  
		sum.len = 0;  
		int num = 0;  
		for (int i = 0; i < len; i++) {  
			num = num * 10 + s[i];  
			sum.s[sum.len++] = num / k;  
			num = num % k;  
		}  
		return sum;  
	}

	int operator % (int k){  
		int sum = 0;  
		for (int i = 0; i < len; i++){  
			sum = sum * 10 + s[i];  
			sum = sum % k;  
		}  
		return sum;  
	} 

	bool operator < (const bign& b) const {
		if (len != b.len)
			return len < b.len;
		for (int i = 0; i < len; i++)
			if (s[i] != b.s[i])
				return s[i] < b.s[i];
		return false;
	}
	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); }
};

int main () {
	bign z = 0;
	int cas, n, m, x;
	scanf("%d", &cas);
	while (cas--) {
		scanf("%d%d", &m, &n);
		bign t = m;
		bign ans = 1;
		for (int i = 1; i < n; i++)
			ans = ans * t;
		for (x = n; x >= 1; x--) if (n%x == 0) {
			t = ans%x;
			if (t == z) break;
		}
		ans = ans / x;
		printf("%d/", n/x);
		ans.put();
		cout << endl;
	}
	return 0;
}