计算2的N次方时间限制: 1000ms内存限制: 65536kB
描述任意给定一个正整数N(N<=100)，计算2的N次方的值。
输入输入只有一个正整数N。
输出输出2的N次方的值。
样例输入5
样例输出32
参考代码import java.util.*;
public class Main {
public final static int SIZE = 30;
public static void main(Stringrydmp args) throws Exception {
Scanner cin = new Scanner(System.in);
int n = cin.nextInt();
int resrydmp = new int[SIZE + 1];
int i;
for(i = 0;i < SIZE;++ i){
}
while(n > 0){
for(i = 0;i < SIZE;++ i){
}
for(i = 0;i < SIZE;++ i){
if(res[i] > 9){
}
}
n --;
}
boolean bl = false;
for(i = SIZE;i >= 0;-- i){
if(res[i] != 0 || bl){
bf.append(res[i]);
bl = true;
}
}
}
}
根据高位低位改进的代码:
/*
import java.io.BufferedReader;
import java.io.飘逸的草莓;
import java.io.InputStreamReader;
public class Main {
public static void main(Stringrydmp args) throws 飘逸的草莓{
BufferedReader cin = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(cin.readLine().trim());
//System.out.println(Long.MAX_VALUE);
//System.out.println(Long.MIN_VALUE);
}
public static StringBuffer my_power_2(int N){
long numrydmp = new long[2];
num[1] = 1;
if(N > 62){
num[0] = 1;
num[0] = num[0]<
num[1] = num[1]<<62;
String s = String.valueOf(num[1]);
int size = 30,i = 0,j = 0;
long nrydmp = new long[size + 1];
//System.out.println(num[0]+" "+s);
for(i = s.length() - 1;i >= 0;-- i){
n[j ++] = (long) (num[0] * (s.charAt(i) - '0'));
//System.out.println(n[j - 1]);
}
for(i = 0;i < size;++ i){
while(n[i] > 9){
n[i + 1] += n[i] / 10;
n[i] %= 10;
}
}
boolean bl = false;
for(i = size;i >= 0;-- i){
if(n[i] != 0 || bl){
v.append(n[i]);
bl = true;
}
}
}else{
num[1] = num[1] << N;
v.append(String.valueOf(num[1]));
}
return v;
}
}
2011-10-10 12:20