Problem 11 of What is the greatest product of four adjacent numbers

本文探讨了一个20x20数字网格中四个相邻数的最大乘积问题,这些数可以沿任何方向(水平、垂直或对角线)排列。通过编程方法解决了这一挑战,并给出了最大乘积的具体数值。

In the 20×20 grid below, four numbers along a diagonal line have been marked in red.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 × 63 × 78 × 14 = 1788696.

What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 20×20 grid?


public class GridProduct {
	public static void main(String args[]){
		int [][] a={
				{8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91,8},
				{49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00},
				{81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65},
				{52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91},
				{22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},
				{24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50},
				{32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70},
				{67,26,20,68,02,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21},
				{24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},
				{21,36,23,9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95},
				{78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14,9,53,56,92},
				{16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57},
				{86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58},
				{19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40},
				{04,52,8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66},
				{88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69},
				{04,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36},
				{20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16},
				{20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54},
				{01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48},
		};
		int mul1=1,mul2=1,mul3=1,mul4=1;
		//down
		for(int i=0;i<20;i++){
			for(int j=0;j<17;j++){
				int pro1=a[i][j]*a[i][j+1]*a[i][j+2]*a[i][j+3];
				if(mul1<pro1){
					mul1=pro1;
				}
			}
		}
		//left
		for(int i=0;i<17;i++){
			for(int j=0;j<20;j++){
				int pro2=a[i][j]*a[i+1][j]*a[i+2][j]*a[i+3][j];
				if(mul2<pro2){
					mul2=pro2;
				}
			}
		}
		//diagonally right
		for(int i=0;i<17;i++){
			for(int j=0;j<17;j++){
				int pro3=a[i][j]*a[i+1][j+1]*a[i+2][j+2]*a[i+3][j+3];
				if(mul3<pro3){
					mul3=pro3;
				}
			}
		}
		//diagonally left
		for(int i=19;i>=3;i--){
			for(int j=0;j<17;j++){
				int pro4=a[i][j]*a[i-1][j+1]*a[i-2][j+2]*a[i-3][j+3];
				if(mul4<pro4){
					mul4=pro4;
				}
			}
		}
		
		System.out.println("mul1="+mul1+",mul2="+mul2+",mul3="+mul3+",mul4="+mul4);
	}
}

Answer:
70600674




import java.util.HashMap;

public class Recss {

	static HashMap<Integer, String> horHashmap=new HashMap<Integer, String>();
	public static int horizental(int [][] array){
		int result=0;
		for(int m=0;m<20;m++){
			for(int n=0;n<17;n++){
				
					int temp=array[m][n]*array[m][n+1]*array[m][n+2]*array[m][n+3];
					if(temp>result){
						horHashmap.put(temp,String.valueOf(array[m][n])+","+String.valueOf(array[m][n+1])+","+String.valueOf(array[m][n+2])+","+String.valueOf(array[m][n+3]));
						result=temp;
					
				}
			}
		}
		return result;
	}
	
	static HashMap<Integer, String> verHashmap=new HashMap<Integer, String>();
	public static int vertal(int [][] array){
		int result=0;
		for(int m=0;m<17;m++){
			for(int n=0;n<20;n++){
					int temp=array[m][n]*array[m+1][n]*array[m+2][n]*array[m+3][n];
					if(temp>result){
						verHashmap.put(temp,String.valueOf(array[m][n])+","+String.valueOf(array[m+1][n])+","+String.valueOf(array[m+2][n])+","+String.valueOf(array[m+3][n]));
						result=temp;
					
				}
			}
		}
		return result;
	}
	static HashMap<Integer, String> xieHashmap=new HashMap<Integer, String>();
	public static int xie(int [][] array){
		int result=0;
		for(int m=0;m<17;m++){
			for(int n=0;n<17;n++){
					int temp=array[m][n]*array[m+1][n+1]*array[m+2][n+2]*array[m+3][n+3];
					if(temp>result){
						xieHashmap.put(temp,String.valueOf(array[m][n])+","+String.valueOf(array[m+1][n+2])+","+String.valueOf(array[m+2][n+3])+","+String.valueOf(array[m+3][n+4]));
						result=temp;
					
				}
			}
		}
		return result;
	}
	static HashMap<Integer, String> fanxiangHashmap=new HashMap<Integer, String>();
	public static int fanxiang(int [][] array){
		int result=0;
		for(int m=19;m>2;m--){
			for(int n=0;n<17;n++){
					int temp=array[m][n]*array[m-1][n+1]*array[m-2][n+2]*array[m-3][n+3];
					if(temp>result){
						fanxiangHashmap.put(temp,String.valueOf(array[m][n])+","+String.valueOf(array[m-1][n+1])+","+String.valueOf(array[m-2][n+2])+","+String.valueOf(array[m-3][n+3]));
						result=temp;
					
				}
			}
		}
		return result;
	}
	public static void main(String[] args) {
		int [][] array={{8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91,8},
						{49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00},
						{81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65},
						{52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91},
						{22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},
						{24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50},
						{32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70},
						{67,26,20,68,02,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21},
						{24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},
						{21,36,23,9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95},
						{78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14,9,53,56,92},
						{16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57},
						{86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58},
						{19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40},
						{04,52,8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66},
						{88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69},
						{04,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36},
						{20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16},
						{20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54},
						{01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48}};
		Recss rec=new Recss();
		System.out.println(horizental(array));
		System.out.println(horHashmap.get(horizental(array)));
		System.out.println(vertal(array));
		System.out.println(verHashmap.get(vertal(array)));
		System.out.println(xie(array));
		System.out.println(xieHashmap.get(xie(array)));
		System.out.println(fanxiang(array));
		System.out.println(fanxiangHashmap.get(fanxiang(array)));
	}
	
}


内容概要:本文介绍了一种基于蒙特卡洛模拟和拉格朗日优化方法的电动汽车充电站有序充电调度策略,重点针对分时电价机制下的分散式优化问题。通过Matlab代码实现,构建了考虑用户充电需求、电网负荷平衡及电价波动的数学模【电动汽车充电站有序充电调度的分散式优化】基于蒙特卡诺和拉格朗日的电动汽车优化调度(分时电价调度)(Matlab代码实现)型,采用拉格朗日乘子法处理约束条件,结合蒙特卡洛方法模拟大量电动汽车的随机充电行为,实现对充电功率和时间的优化分配,旨在降低用户充电成本、平抑电网峰谷差并提升充电站运营效率。该方法体现了智能优化算法在电力系统调度中的实际应用价值。; 适合人群:具备一定电力系统基础知识和Matlab编程能力的研究生、科研人员及从事新能源汽车、智能电网相关领域的工程技术人员。; 使用场景及目标:①研究电动汽车有序充电调度策略的设计与仿真;②学习蒙特卡洛模拟与拉格朗日优化在能源系统中的联合应用;③掌握基于分时电价的需求响应优化建模方法;④为微电网、充电站运营管理提供技术支持和决策参考。; 阅读建议:建议读者结合Matlab代码深入理解算法实现细节,重点关注目标函数构建、约束条件处理及优化求解过程,可尝试调整参数设置以观察不同场景下的调度效果,进一步拓展至多目标优化或多类型负荷协调调度的研究。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值