C#练习题答案: 产品和LCMS之间的差异总和【难度：1级】--景越C#经典编程题库,1000道C#基础练习题等你来挑战
2024-06-08 13:56:37
产品和LCMS之间的差异总和【难度:1级】:
答案1:
using System.Linq;public class Kata {static int gcd(int a, int b) {if(a == 0) return b;if(b == 0) return a;return gcd(b, a % b);}static int lcm(int a, int b) {if(a == 0 || b == 0) return 0;return a * b / gcd(a, b);}public static int SumDifferencesBetweenProductsAndLCMs(int[][] pairs) {return pairs.Select(p => p[0] * p[1] - lcm(p[0], p[1])).Sum();}
}答案2:
using System.Linq;
public class Kata
{public static int SumDifferencesBetweenProductsAndLCMs(int[][] pairs){return Enumerable.Range(0, pairs.Length).Select(x => pairs[x][1] == 0 ? 0 : (pairs[x][0] * pairs[x][1]) - GetLCM(pairs[x][0], pairs[x][1])).Sum();}static int gcf(int a, int b){while (b != 0){int temp = b;b = a % b;a = temp;}return a;}static int GetLCM(int a, int b) => (a / gcf(a, b)) * b;
}答案3:
using System;
using System.Collections.Generic;public class Kata
{public static int GetResultFromMultiArray(int[] array){int a,b,c,remainder,total;a = array[0];b = array[1];c = a * b;total = a * b;while (b != 0){remainder = a % b;a = b;b = remainder; }if(a == 0 ){return 0;}c = c / a;total = total - c;return total;}public static int SumDifferencesBetweenProductsAndLCMs(int[][] pairs){List<int> Product = new List<int>();List<int> LCM = new List<int>();for (var i = 0; i < pairs.GetLength(0); i++){var Num = GetResultFromMultiArray(pairs[i]);Product.Add(Num);}int TotalSum = 0;for (var i = 0; i < Product.Count; i++){TotalSum += Product[i];}Console.WriteLine("[{0}] \n The total sum is {1}", string.Join(", ", Product), TotalSum);return TotalSum;}
}答案4:
public class Kata
{public static int SumDifferencesBetweenProductsAndLCMs(int[][] pairs){int multiply = 0;int lcm = 0;int res = 0;foreach (int[] row in pairs){if (row[1] != 0){var i = 0;multiply = row[0] * row[1];do{i++;}while ((i * row[0]) % row[1] != 0);lcm = i * row[0];res += (multiply - lcm);}}return res;}
}答案5:
using System;public class Kata
{public static int SumDifferencesBetweenProductsAndLCMs(int[][] pairs){int[] arrayOfProducts = new int[pairs.GetLength(0)];int[] arrayOfLCM = new int[pairs.GetLength(0)];int result = 0;for (int i = 0; i < pairs.GetLength(0); ++i){arrayOfProducts[i] = pairs[i][0] * pairs[i][1];arrayOfLCM[i] = calculateLCM(pairs[i][0], pairs[i][1]);}for (int i = 0; i < pairs.GetLength(0); ++i){result += arrayOfProducts[i] - arrayOfLCM[i];}return result;}public static int calculateLCM(int x, int y){if (x == 0 || y == 0) return 0;return Math.Abs(x * y) / calculateNOD(x, y);}public static int calculateNOD(int x, int y){while (x != y){if (x > y)x = x - y;elsey = y - x;}return x;}
}答案6:
using System;public class Kata {public static int SumDifferencesBetweenProductsAndLCMs( int[][] pairs ) {var sum = 0;foreach ( var pair in pairs ) {long a = pair [ 0 ];long b = pair [ 1 ];var diff = a*b - Lcm( a, b );sum += ( int ) diff;}return sum;}private static long Lcm( long a, long b ) {var gcd = Gcd( a, b );return ( a*b )/Math.Max( gcd, 1 );}private static long Gcd( long a, long b ) {return b == 0 ? a : Gcd( b, a%b );}
}答案7:
public class Kata
{public static int SumDifferencesBetweenProductsAndLCMs(int[][] pairs){int result = 0;for (int i = 0; i < pairs.Length; i++){result += (pairs[i][0] * pairs[i][1]) - LCM(pairs[i][0], pairs[i][1]);}return result;}static int GCD (int a, int b){ if (a == 0)return b;if (b == 0)return a;return GCD(b, a % b);}static int LCM (int a, int b){if (a == 0 || b == 0)return 0;return (a * b) / GCD(a, b);}
}答案8:
using System.Linq;public class Kata
{public static int SumDifferencesBetweenProductsAndLCMs(int[][] pairs){int sum = 0;for (int i=0; i<pairs.Length; i++){int product = (pairs[i][0] * pairs[i][1]);sum += product - (product / GCD(pairs[i][0], pairs[i][1]));}return sum;}public static int GCD(int a, int b){if (a + b == 0)return 1;else if (b == 0)return a;return GCD(b, a%b);//}
}答案9:
using System;public class Kata
{public static int SumDifferencesBetweenProductsAndLCMs(int[][] pairs){int[,] arrLCM = new int[pairs.Length, 1];int[,] arrProd = new int[pairs.Length, 1];for (int i = 0; i < pairs.Length; i++){Console.WriteLine(pairs[i][0]);Console.WriteLine(pairs[i][1]);Console.WriteLine();arrProd[i, 0] = pairs[i][0] * pairs[i][1];}for (int i = 0; i < pairs.Length; i++){int LCM = 0;int start;if (pairs[i][0] <= pairs[i][1])start = pairs[i][0];elsestart = pairs[i][1];if (pairs[i][0] == pairs[i][1])LCM = pairs[i][0];elsefor (int j = start + 1; j <= pairs[i][0] * pairs[i][1]; j++){if (j % pairs[i][0] == 0 && j % pairs[i][1] == 0){LCM = j;break;}}arrLCM[i, 0] = LCM;}int ret = 0;for (int i = 0; i < arrLCM.Length; i++){ret += arrProd[i, 0] - arrLCM[i, 0];}return ret;}
}答案10:
using System.Linq;public class Kata
{public static int calc(int i1, int i2) {int sum1 = 0;for (int i = 1 ; i <= i1 * i2; i++){if ((i%i1 ==0) && (i%i2 ==0) ){ sum1 = i;break;}} return sum1;}public static int SumDifferencesBetweenProductsAndLCMs(int[][] pairs){int i = pairs.Select(t => t[0]*t[1] - calc(t[0],t[1]) ).Sum();return i;}
}C#练习题答案: 产品和LCMS之间的差异总和【难度：1级】--景越C#经典编程题库,1000道C#基础练习题等你来挑战相关推荐
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