BigInteger num = BigInteger.valueOf(123456789l); BigInteger result = num; for(int i = 1; i < 100; i++) result = result.multiply(num); System.out.println(result.toString());
在java中那样的题目太少了吧,,
import java.math.BigInteger; class bigInteger { public static void main(String[] args) {
BigInteger num = BigInteger.valueOf(123456789l); num = num.pow(100); System.out.println(num); } }
BigInteger num = BigInteger.valueOf(123456789l);
BigInteger result = num;
for(int i = 1; i < 100; i++)
result = result.multiply(num);
System.out.println(result.toString());
class bigInteger { public static void main(String[] args) {
BigInteger num = BigInteger.valueOf(123456789l); num = num.pow(100); System.out.println(num);
}
}
这么大的数,肯定是用字符串来存,所以比较影响效率的是如何减少从char转换成int的次数