Question

JAVA | 100分求1的1次方+2的2次方+3的3次方+4的4次方+。。。+20的20次方的算法

求详细算法 谢谢！
给予2个以上的算法的 没多加一个 追加50分 谢谢
第一个回答者 的结果错误 是153 第二位朋友希望给出详细算法 第三位朋友看错题目了 不是 20*20 是20的20次方

Answer 1

public class Main {
public static void main (String []args) {
double t=0;
int i;
int j=1;
for(i=1;i<=20;i++){
j=1;
for(int k=1;k<=i;k++){
j=j*i;
}
t+=j;
}
System.out.println(t);
}
}
3.766126E9
给分来

Answer 2

public class test {

public static long sq(long m){
long sum=1;
for(long i=0;i<m;i++){
sum*=m;
}
return sum;
}
public static long add(long n){
long tal=0;
for(long i=1;i<=n;i++){
tal=tal+sq(i);
}
return tal;
}
//---以上为第一种方法
//-------------------------------------------
//一下为第二中方法
public static long sq2(long m){
return (long)java.lang.Math.pow(m,m);
}
public static long add2(long n){
long tal=0;
for(long i=1;i<=n;i++){
tal=tal+sq2(i);
}
return tal;
}

public static void main(String[] args) {
test t=new test();

System.out.println(t.add(4));
System.out.println(t.add2(4));
//第二中方 pow后的值若大于double的临界值 会报错
}
}

Answer 3

//楼主满意的收下吧：

import java.math.BigInteger;

public class Test2 {

public static void main(String[] args) {
Number[] number = new Number[20];
String str = "";
for (int i = 0; i < 20; i++) {
str = String.valueOf(i + 1);
number[i] = new Number(new BigInteger(str), i + 1);
}

BigInteger sum = new BigInteger ("0");

for (int i = 0; i < number.length; i++) {
sum = sum.add(number[i].num.pow(number[i].n));
}

System.out.println(sum);
}
}

class Number {
BigInteger num;
int n;
public Number(BigInteger num, int n) {
this.num = num;
this.n = n;
}
}

Answer 4

Hello guy!
The following code is achieved as enterprise business app,Maybe you
are searching for which is explicitly used numeric analysis...
Just hope a little help for you! Thank you...

import java.math.BigDecimal;

public class BigDecimalTest {

public static BigDecimal function(int index) {
BigDecimal num = new BigDecimal(index);
return num.pow(index);
}

public static void main(String[] args) {

BigDecimal result = new BigDecimal(0);
for (int i = 1; i <= 20; i++) {
result = result.add(function(i));
}

System.out.println("1的1次方+2的2次方+3的3次方+4的4次方....+20的20次方的结果为:"+result.toPlainString());
}
}

Answer 5

public class Jc {
public static void main (String []args) {
System.out.println (jc(5));
}
public static int jc (int i) {
int j=1;
int s=1;
int k=0;
while (j<=i){
s=s*j;
k=s+k;
j++;
}
return k;
}
}