poj3892 高精度

RSA Factorization

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 471		Accepted: 124

Description

The positive integer n is given. It is known that n = p * q, where p and q are primes, q <= p and |q - kp| <= 10 ⁵ for some given positive integer k. You must find p and q.

Input

Each line contains integers n (1 < n < 10 ¹²⁰) and k (0 < k < 10 ⁸).

Output

For each pair of numbers n and k print in separate line the product p * q such that q <= p.

Sample Input

35 1 
121 1 
1000730021 9

Sample Output

5 * 7 
11 * 11 
10007 * 100003

java大数功能很强大啊，有木有！可能，这里，是我认为，java比c++,好用的地方吧！想不通啊，为什么用c++写就非要超时，用java写这么容易就过了，java 是个好东西啊！题意不想理解错了，其实，就是，要把一个n可以分成两个数，把这两个数求出来，而且一定，是只有一个组合，没想到，这题这么坑啊！无语，但是学学java是个好东西啊！

import java.math.BigInteger;
import java.util.Scanner;
public class Main {
    static BigInteger n;
	static BigInteger k;
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);//相当于scanf
		BigInteger one = BigInteger.ONE;
		BigInteger two = BigInteger.valueOf(2);//2要转换
		BigInteger zero = BigInteger.ZERO;
		BigInteger l,r,tmp,mid;
		mid=zero;
		while(cin.hasNext())//相当于！＝ＥＯＦ
		{
			n = cin.nextBigInteger();//相当于scanf
			k = cin.nextBigInteger();
			k = k.multiply(n);
			l = zero;
			r = k;
			while(l.compareTo(r)<=0)
			{
				mid = l.add(r).divide(two);
				tmp = mid.multiply(mid);
				if(tmp.compareTo(k) == 0)
					break;
				if(tmp.compareTo(k) < 0)
					l=mid.add(one);
				else
					r=mid.subtract(one);
			}
            BigInteger p,q;
			p=q= mid;
			while(true)
			{
				if(n.mod(p).equals(zero) && !p.equals(one) && !p.equals(n))
				{
					q = n.divide(p);
					break;
				}
				if(n.mod(q).equals(zero)&& !q.equals(one) && !q.equals(n))
				{
					p = n.divide(q);
					break;
				}
				p = p.add(one);
				q = q.subtract(one);
			}

			if(p.compareTo(q)>0)
			{
				tmp = p;
				p = q;
				q = tmp;
			}
			System.out.println(p+" * "+q);
		}
	}

}
找了一些函数
JAVA大数处理(BigInteger,BigDecimal)

这两个类都在java.math.*包中，因此每次必须在开头处引用该包。
Ⅰ基本函数：
1.valueOf(parament); 将参数转换为制定的类型
比如 int a=3;
BigInteger b=BigInteger.valueOf(a);
则b=3;
String s=”12345”;
BigInteger c=BigInteger.valueOf(s);
则c=12345；
2.add(); 大整数相加
BigInteger a=new BigInteger(“23”);
BigInteger b=new BigInteger(“34”);
a. add(b);
3.subtract(); 相减
4.multiply(); 相乘
5.divide(); 相除取整
6.remainder();取余
7.pow(); a.pow(b)=a^b//b不能为大数
8.gcd(); 最大公约数
9.abs(); 绝对值
10.negate();取反数
11.mod(); a.mod(b)=a%b=a.remainder(b);
13.punlic int comareTo();
14.boolean equals(); 是否相等
15.BigInteger构造函数：
一般用到以下两种：
BigInteger(String val);
将指定字符串转换为十进制表示形式；
BigInteger(String val,int radix);
将指定基数的BigInteger的字符串表示形式转换为BigInteger
Ⅱ.基本常量：
A=BigInteger.ONE 1
B=BigInteger.TEN 10
C=BigInteger.ZERO 0
Ⅲ.基本操作
1. 读入：
用Scanner类定义对象进行控制台读入,Scanner类在java.util.*包中
Scanner cin=new Scanner(System.in);// 读入
while(cin.hasNext()) //等同于!=EOF
{
int n;
BigInteger m;
n=cin.nextInt(); //读入一个int;
m=cin.BigInteger();//读入一个BigInteger;
System.out.print(m.toString());
}
Ⅳ.运用
四则预算：
import java.util.Scanner;
import java.math.*;
import java.text.*;
public class Main 
{
public static void main(String args[])
{
Scanner cin = new Scanner ( System.in );
BigInteger a,b;
int c;
char op;
String s;

while( cin.hasNext() )
{
a = cin.nextBigInteger();
s = cin.next();
op = s.charAt(0);
if( op == '+')
{
b = cin.nextBigInteger();
System.out.println(a.add(b));
}
else if( op == '-')
{
b = cin.nextBigInteger();
System.out.println(a.subtract(b));
}
else if( op == '*')
{
b = cin.nextBigInteger();
System.out.println(a.multiply(b));
}
else
{
BigDecimal a1,b1,eps;
String s1,s2,temp;
s1 = a.toString();
a1 = new BigDecimal(s1);
b = cin.nextBigInteger();
s2 = b.toString();
b1 = new BigDecimal(s2);
c = cin.nextInt();
eps = a1.divide(b1,c,4);
//System.out.println(a + " " + b + " " + c);
//System.out.println(a1.doubleValue() + " " + b1.doubleValue() + " " + c);
System.out.print( a.divide(b) + " " + a.mod(b) + " ");
if( c != 0)
{
temp = "0.";
for(int i = 0; i < c; i ++) temp += "0";
DecimalFormat gd = new DecimalFormat(temp);
System.out.println(gd.format(eps));
}
else System.out.println(eps);
}
}
}
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
    
public static void main(String[] args) {
BigInteger n;
BigInteger k;
BigInteger c;
Scanner cin = new Scanner(System.in);//相当于scanf
BigInteger one = BigInteger.ONE;
BigInteger two = BigInteger.valueOf(2);//2要转换
BigInteger zero = BigInteger.ZERO;
BigInteger l,r,tmp,mid;
mid=zero;
double a;
while(cin.hasNext())//相当于！＝ＥＯＦ
{
n = cin.nextBigInteger();//相当于scanf
k = cin.nextBigInteger();
a = cin.nextDouble();
System.out.printf("%.3f\n",a);//double　float 都要用f而不是lf
System.out.println(n+""+k);
c=n.add(k);
System.out.println("add"+c);
c=n.subtract(k);
System.out.println("substrack"+c);
c=n.multiply(k);
System.out.println("multiply"+c);
c=n.divide(k);
System.out.println("divide"+c);
c=n.remainder(k);
System.out.println("remainder"+c);
c=n.pow(2);//这里不能是大数
System.out.println("pow"+c);
c=n.gcd(k);
System.out.println("gcd"+c);
c=n.abs();
System.out.println("abs"+c);
c=n.negate();
System.out.println("negate"+c);
System.out.printf("compare%d\n",n.compareTo(k));
//System.out.printf("equal%d\n", n.equals(k));
String mm="23232";


}
}


}