Question

编写java程序，double tind(int n) 方法功能求2的N次方 程序中调用该方法 输出2的40次方并输出

Answer 1

public class test10 {

/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
double b;
b=tind(40);
System.out.println(b);

}
static double tind(int n) {
double s=1;
for(int i=0;i<n;i++){
s=2*s;
}
return s;
}

}

Answer 2

import java.util.Vector;

public class Test {//a的b次幂

public static void main(String[] args) {

int m=40;
Test test = new Test();
double pow = test.tind(m);
System.out.println(pow);
}

private static double tind(int m) {
int NUM = 2;
Vector<Integer> myVector = new Vector<Integer>();
int bigNum = 1;
myVector.add(bigNum);
int i;
int certerNum = 0;
int newNum;
for (int j = 0; j < m; j++) {
int flag = 0;
for (i = 0; i < myVector.size(); i++) {
certerNum = ((Integer) myVector.elementAt(i)) * NUM + flag;
newNum = certerNum % 10;
flag = certerNum / 10;
myVector.set(i, newNum);
}
if ((certerNum >= 10) && (i == myVector.size())) {
myVector.add(flag);
}
}
String str = "";
for (int k = myVector.size() - 1; k >= 0; k--) {
str += myVector.elementAt(k);
}
System.out.println(str);
double d = Double.parseDouble(str);
return d;
}
}

Answer 3

计算 2的多少次方 直接用 移位操纵 效率最高

例如 2<<1 =2^2
2<<2=2^3
double tind(int n){
if(n=0){
return 1;
}
return 2<<(n-1)
}

Answer 4

double tind(int n){
System.out.println(Math.pow(2, 40));
return Math.pow(2, 40);
}