标签：练习题 数字 C# List System long new using public
取一个数字并将它的数字总和提升到连续的力量和… Eureka !!【难度:2级】:
答案1:
using System;
using System.Collections.Generic;
using System.Linq;
public class SumDigPower {
public static long[] SumDigPow(long a, long b)
{
List<long> values = new List<long>();
for (long x = a; x <= b; x++)
{
if (x.ToString().Select((c, i) => Math.Pow(Char.GetNumericValue(c), i + 1)).Sum() == x)
values.Add(x);
}
return values.ToArray();
}
}
答案2:
using System;
using System.Linq;
using System.Collections.Generic;
public class SumDigPower {
public static long[] SumDigPow(long a, long b)
{
List<long> r = new List<long>();
for(long l = a; l<=b; l++){
int n = 1;
long sum = 0;
foreach(char c in (""+l)){
sum+=(long)Math.Pow(Int64.Parse(""+c), n);
n++;
}
if(sum==l){
r.Add(l);
}
}
return r.ToArray();
}
}
答案3:
using System;
using System.Collections.Generic;
public class SumDigPower {
public static long[] SumDigPow(long a, long b)
{
if (b < a)
return new long[0];
List<long> result = new List<long>();
for (long i = a; i <= b; i++)
{
if (IsEureka(i))
result.Add(i);
}
return result.ToArray();
}
public static bool IsEureka(long num)
{
int digitCount = 0;
long i = num;
while (i > 0)
{
digitCount++;
i /= 10;
}
if (digitCount == 1)
return true;
long sum = 0;
i = num;
while (i > 0)
{
sum += (long)Math.Pow(i % 10, digitCount--);
if (sum > num)
return false;
i /= 10;
}
return sum == num;
}
}
答案4:
using System;
using System.Collections.Generic;
using System.Linq;
public class SumDigPower
{
public static long[] SumDigPow(long a, long b)
{
List<long> values = new List<long>();
for(; a <= b; a++)
{
var digits = a.ToString().Select(x => int.Parse(x.ToString()));
long rslt = 0;
int itrs = 1;
foreach(var digit in digits)
{
rslt += (long)Math.Pow(digit, itrs++);
}
if (rslt == a)
values.Add(rslt);
}
return values.ToArray();
}
}
答案5:
using System;
using System.Linq;
using System.Collections.Generic;
public class SumDigPower {
public static long[] SumDigPow(long a, long b)
{
List<long> ls = new List<long>();
for (long n = a; n <= b; n++) //remember <= since it's inclusive
{
var digitsPow = n.ToString() //seperate digits
.Select(x => x-'0') //convert chars into ints
.Select((x, i) => Math.Pow(x, i+1)) //raise ints to power of digit position
.Sum();
if (digitsPow == n) ls.Add(n);
}
return ls.ToArray();
}
}
答案6:
using System;
public class SumDigPower {
public static long[] SumDigPow(long a, long b)
{
long[] result={};
int rL=0;
for(long i=a;i<=b;i++)
{
var S = i.ToString ().ToCharArray ();
long I = 0;
for(int n=0;n<S.Length;n++){
int Sn = S [n]-48;
I = I + (long)(Math.Pow(Sn,n+1));
if (I > i) {
break;
}
}
if (I == i) {
Array.Resize (ref result, ++rL);
result [rL-1] = i;
}
}
return result;
}
}
答案7:
using System.Collections.Generic;
using System.Linq;
using System;
public class SumDigPower {
public static long[] SumDigPow(long a, long b)
{
List<long> found = new List<long>();
List<long> numbers = new List<long>();
for (long l = a; l <= b; l++)
numbers.Add(l);
numbers.AsParallel().ForAll(num =>
//numbers.ForEach(num =>
{
long temp = num;
double sum = 0;
List<int> digits = new List<int>();
while (temp / 10 > 0)
{
digits.Add((int)(temp % 10));
temp /= 10;
}
digits.Add((int)(temp % 10));
for (int l = 0; l < digits.Count; l++)
sum += Math.Pow(digits[l], digits.Count - l);
if (sum == num)
lock (found)
found.Add(num);
});
return found.ToArray();
}
}
答案8:
using System.Collections.Generic;
using System;
public class SumDigPower {
public static long[] SumDigPow(long a, long b)
{
var result = new List<long>();
for (long i = a; i < b; i++)
{
double temp = 0;
var arrayStringNum = i.ToString().ToCharArray();
for (int j = 0; j < arrayStringNum.Length; j++)
{
var tempNum = int.Parse(arrayStringNum[j].ToString());
temp += Math.Pow(tempNum, j + 1);
}
if (temp == i)
result.Add(i);
}
return result.ToArray();
}
}
答案9:
using System;
using System.Linq;
using System.Collections.Generic;
public class SumDigPower {
public static long[] SumDigPow(long a, long b)
{
if(a > b)
return new long[1];;
var list = new List<long>();
foreach(var number in Enumerable.Range((int)a, (int)b-(int)a))
{
var numbers = $"{number}".ToCharArray().Select(p => int.Parse($"{p}"));
int powerIndex = 0;
var sum = numbers.Sum(p => Math.Pow(p, ++powerIndex));
if(sum == number)
list.Add(Convert.ToInt64(number));
}
return list.ToArray<long>();
}
}
答案10:
using System;
using System.Collections.Generic;
using System.Linq;
public class SumDigPower {
public static long[] SumDigPow(long a, long b)
{
long start = a > b ? b : a;
long end = a > b ? a : b;
List<long> eurekas = new List<long>();
for (long i = start; i <= end; i++) {
if (isEureka(i))
eurekas.Add(i);
}
return eurekas.ToArray();
}
private static bool isEureka(long n) {
char[] digits = n.ToString().ToCharArray();
long eureka = 0;
for (int i = 0; i < digits.Length; i++) {
eureka += (long)Math.Pow(Convert.ToDouble(digits[i].ToString()), i+1);
}
return n == eureka;
}
}
标签：练习题,数字,C#,List,System,long,new,using,public 来源： https://blog.csdn.net/weixin_45444821/article/details/100846183
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