JAVA处理大数入门 转载
在用C或者C++处理大数时感觉非常麻烦,但是在JAVA中有两个类BigInteger和BigDecimal分别表示大整数类和大浮点数类,
两个类的对象能表示最大范围理论上能够表示无线大的数,只要计算机内存足够大。 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
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);
12.max(); min();
13.public 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.nextBigInteger();//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.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);
}
}
}
}
例子:求一个n元素集合中k个元素的组合数量:C(n,k) = n!/(k! * (n-k)!)
import java.math.BigInteger;
import java.util.*;
import java.io.*;
public class FactorialBigIntegerTest {
public static void main(String[] args) throws Exception {
Scanner cin=new Scanner(System.in);
int T = cin.nextInt();
for(int i = 0;i < T;i++){
int n=cin.nextInt(),k=cin.nextInt();
BigInteger nf= new BigInteger(""+factorial(n));
BigInteger kf= new BigInteger(""+factorial(k));
BigInteger nk = new BigInteger(""+factorial(n-k));
BigInteger n1 = new BigInteger(""+nf.divide(kf));
System.out.println(n1.divide(nk));
}
}
private static final BigInteger factorial(int n){
BigInteger bigInteger=new BigInteger(""+n);
if(n==0)
return BigInteger.ONE;
else
return factorial(n-1).multiply(bigInteger);
}
}
输入 2000 50
运行结果:
19961695734279236211032903597262781251004330223561192760088208846391036329722665786642280715824928160
posted on 2011-08-13 20:50 more think, more gains 阅读(525) 评论(0) 编辑 收藏 举报