// By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.// // 3// 7 4// 2 4 6// 8 5 9 3// // That is, 3 + 7 + 4 + 9 = 23.// // Find the maximum total from top to bottom of the triangle below:// // 75// 95 64// 17 47 82// 18 35 87 10// 20 04 82 47 65// 19 01 23 75 03 34// 88 02 77 73 07 63 67// 99 65 04 28 06 16 70 92// 41 41 26 56 83 40 80 70 33// 41 48 72 33 47 32 37 16 94 29// 53 71 44 65 25 43 91 52 97 51 14// 70 11 33 28 77 73 17 78 39 68 17 57// 91 71 52 38 17 14 91 43 58 50 27 29 48// 63 66 04 68 89 53 67 30 73 16 69 87 40 31// 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23// // NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)using System;using System.Collections.Generic;using System.Text;namespace projecteuler018{ class Program { static void Main(string[] args) { F2(); } #region F1() private static void F1() { Console.WriteLine(new System.Diagnostics.StackTrace().GetFrame(0).GetMethod()); DateTime timeStart = DateTime.Now; int[] data = new int[] { 75, 95, 64, 17, 47, 82, 18, 35, 87, 10, 20, 04, 82, 47, 65, 19, 01, 23, 75, 03, 34, 88, 02, 77, 73, 07, 63, 67, 99, 65, 04, 28, 06, 16, 70, 92, 41, 41, 26, 56, 83, 40, 80, 70, 33, 41, 48, 72, 33, 47, 32, 37, 16, 94, 29, 53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14, 70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57, 91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48, 63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31, 04, 62, 98, 27, 23, 09, 70, 98, 73, 93, 38, 53, 60, 04, 23 }; Console.WriteLine(largestRoute1(data)); Console.WriteLine("Total Milliseconds is " + DateTime.Now.Subtract(timeStart).TotalMilliseconds + "\n\n"); } private static int largestRoute1(int[] data) { if (data.Length == 1) { return data[0]; } //获取层数 int layer = 1; for (int i = data.Length; ; layer++) { i -= layer; if (i == 0) { break; } } //获取左下三角 //012345 -> 134 int[] leftData = new int[data.Length - layer]; int skipCount = 0; int currentCount = 0; int currentPosition = 0; for (int i = 0; i < data.Length; i++) { if (currentCount-- > 0) { leftData[currentPosition++] = data[i]; } else { skipCount++; currentCount = skipCount; } } //获取右下三角 //012345 -> 245 int[] rightData = new int[data.Length - layer]; skipCount = -1; currentCount = 0; currentPosition = 0; for (int i = 0; i < data.Length; i++) { if (currentCount-- > 0) { rightData[currentPosition++] = data[i]; } else { skipCount++; currentCount = skipCount; } } return Math.Max(largestRoute1(leftData), largestRoute1(rightData)) + data[0]; } #endregion F1() #region F2() /// <summary> /// 每个索引节点的最大路径值 /// </summary> static int[] route; /// <summary> /// 数据总共的层数 /// </summary> static int totalLayer; private static void F2() { Console.WriteLine(new System.Diagnostics.StackTrace().GetFrame(0).GetMethod()); DateTime timeStart = DateTime.Now; int[] data = new int[] { 75, 95, 64, 17, 47, 82, 18, 35, 87, 10, 20, 04, 82, 47, 65, 19, 01, 23, 75, 03, 34, 88, 02, 77, 73, 07, 63, 67, 99, 65, 04, 28, 06, 16, 70, 92, 41, 41, 26, 56, 83, 40, 80, 70, 33, 41, 48, 72, 33, 47, 32, 37, 16, 94, 29, 53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14, 70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57, 91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48, 63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31, 04, 62, 98, 27, 23, 09, 70, 98, 73, 93, 38, 53, 60, 04, 23 }; route = new int[data.Length]; Console.WriteLine(largestRoute2(data, 0, 0)); Console.WriteLine("Total Milliseconds is " + DateTime.Now.Subtract(timeStart).TotalMilliseconds + "\n\n"); } /// <summary> /// 获取某个三角的最大路径和 /// </summary> /// <param name="data">三角数据</param> /// <param name="layer">层数，从1开始</param> /// <param name="index">在数组中的索引，从0开始，左下三角索引为index+currentlayer，右下三角的索引是index+currentlayer+1</param> /// <returns></returns> private static int largestRoute2(int[] data, int layer, int index) { if (layer == 1) { route[index] = data[0]; return data[0]; } if (route[index] != 0 && layer != 0) { return route[index]; } //如果是0层，则获取层数 if (layer == 0) { layer = 1; for (int i = data.Length; ; layer++) { i -= layer; if (i == 0) { break; } } totalLayer = layer; } int currentLayer = totalLayer - layer + 1; //获取左下三角 //012345 -> 134 int[] leftData = new int[data.Length - layer]; int skipCount = 0; int currentCount = 0; int currentPosition = 0; for (int i = 0; i < data.Length; i++) { if (currentCount-- > 0) { leftData[currentPosition++] = data[i]; } else { skipCount++; currentCount = skipCount; } } //获取右下三角 //012345 -> 245 int[] rightData = new int[data.Length - layer]; skipCount = -1; currentCount = 0; currentPosition = 0; for (int i = 0; i < data.Length; i++) { if (currentCount-- > 0) { rightData[currentPosition++] = data[i]; } else { skipCount++; currentCount = skipCount; } } route[index] = Math.Max(largestRoute2(leftData, layer - 1, index + currentLayer), largestRoute2(rightData, layer - 1, index + currentLayer + 1)) + data[0]; return route[index]; } #endregion F2() }}/*Void F1()1074Total Milliseconds is 61.0077Void F2()1074Total Milliseconds is 7.0009 By GodMoon*/
