【2014 Multi-University Training Contest 2 1002】/【HDU 4873】 ZCC Loves Intersection

果然，或滥用零件，啥都不说了。我们欣慰地学习阅读。这两天残疾儿童是数学。

这是求所需的问题，不明确。贴上官方的解题报告。

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留着慢慢研究    。

好吧看到其它人写的发现有自带函数。就再贴一个新的。

就得在最以下：

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
	static BigInteger [][] c = new BigInteger[110][110];
	public static void del() {
		for(int i = 0; i <= 105; i ++)
			c[i][0] = c[i][i] = BigInteger.ONE;
		for(int i = 1; i <= 105; i ++)
		{
			for(int j = 1; j < i; j ++)
				c[i][j] = (c[i-1][j-1] .add(c[i-1][j]));
		}
	}
	public static void main(String[] args)  {
		Scanner cin = new Scanner(System.in);
		int d;
		BigInteger n=BigInteger.ZERO,up=BigInteger.ZERO,down=BigInteger.ZERO,temp=BigInteger.ZERO;
		del();
		while(cin.hasNext())
		{
			n=cin.nextBigInteger();
			d=cin.nextInt();
			up=c[d][2].multiply(n.add(BigInteger.valueOf(4)).pow(2));
			down=BigInteger.valueOf(9).multiply(n.pow(d));
			if(up.compareTo(down)==0)
			{
				System.out.println(1);
			}
			else {
				temp=up.gcd(down);
				System.out.println(up.divide(temp)+"/"+down.divide(temp));
			}
		}
		cin.close();
	}
}

以下是代码（旧）：

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
	static BigInteger [][] c = new BigInteger[110][110];
	public static void del() {
		for(int i = 0; i <= 105; i ++)
			c[i][0] = c[i][i] = BigInteger.ONE;
		for(int i = 1; i <= 105; i ++)
		{
			for(int j = 1; j < i; j ++)
				c[i][j] = (c[i-1][j-1] .add(c[i-1][j]));
		}
	}
	public static BigInteger gcd(BigInteger a ,BigInteger b) {
		if(a .compareTo(b)<0)
			return gcd(b,a);
		if(a .mod(b).compareTo(BigInteger.ZERO)== 0)
			return b;
		return gcd(b, a.mod(b));
	}
	public static BigInteger pow(BigInteger a ,int b) {
		BigInteger ans = BigInteger.ONE;
		for(int i=1;i<=b;i++)
		{
			ans=ans.multiply(a);
		}
		return ans;
	}
	public static void main(String[] args)  {
		Scanner cin = new Scanner(System.in);
		int d;
		BigInteger n=BigInteger.ZERO,up=BigInteger.ZERO,down=BigInteger.ZERO,temp=BigInteger.ZERO;
		del();
		while(cin.hasNext())
		{
			n=cin.nextBigInteger();
			d=cin.nextInt();
			up=c[d][2].multiply(pow(n.add(BigInteger.valueOf(4)),2));
			down=BigInteger.valueOf(9).multiply(pow(n, d));
			if(up.compareTo(down)==0)
			{
				System.out.println(1);
			}
			else {
				temp=gcd(up, down);
				System.out.println(up.divide(temp)+"/"+down.divide(temp));
			}
		}
		cin.close();
	}
}

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