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  • 以下这个20×20网格四个处于同一对角线上相邻数字用粗体标了出来: 这四个数字乘积是:26×63×78×14=178869626. 在这个20×20网格,处于任何方向上(上,下,左,右或者对角线)的四个相邻数字...

    题目

    在以下这个20×20的网格中,四个处于同一对角线上的相邻数字用粗体标了出来:
    这四个数字的乘积是:26×63×78×14=1788696.
    在这个20×20网格中,处于任何方向上(上,下,左,右或者对角线)的四个相邻数字的乘积的最大值是多少?

    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

    解题方法

    一共有4个方向的连线,横(—),竖(|),左对角线(\),右对角线(/)。
    依次计算四个方向的最大值,取最大值。

    程序

    编程的时候需要注意一点,上面的20×20的网格中,为了美观,所有的数字都是两位的,不足两位前面补0。但是这带来一个问题,编程语言中以0开始的直接数表示这个数是八进制表示的。如012=10,但是08是非法的八进制数。因为八进制只有0~7这8个数。所以,在输入数据时要注意这点。顺便一提,以0x开始的数是十六进制数,如0x12=18。

    public static void solve() {
        int[][] num = { { 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 max = 0;
        // 横向,从左到右
        for (int i = 0; i < 20; i++) {
            for (int j = 0; j < 17; j++) {
                int multi = 1;
                for (int k = 0; k < 4; k++) {
                    multi *= num[i][j + k];
                }
                max = multi > max ? multi : max;
            }
        }
        // 竖向,从上到下
        for (int i = 0; i < 17; i++) {
            for (int j = 0; j < 20; j++) {
                int multi = 1;
                for (int k = 0; k < 4; k++) {
                    multi *= num[i + k][j];
                }
                max = multi > max ? multi : max;
            }
        }
        // 右对角线,从左下到右上
        for (int i = 3; i < 20; i++) {
            for (int j = 0; j < 17; j++) {
                int multi = 1;
                for (int k = 0; k < 4; k++) {
                    multi *= num[i - k][j + k];
                }
                max = multi > max ? multi : max;
            }
        }
        // 左对角线,从左上到右下
        for (int i = 0; i < 17; i++) {
            for (int j = 0; j < 17; j++) {
                int multi = 1;
                for (int k = 0; k < 4; k++) {
                    multi *= num[i + k][j + k];
                }
                max = multi > max ? multi : max;
            }
        }
        // 输出最大值
        System.out.println(max);
    }
    展开全文
  • //begin==end说明数组为奇数 if(begin==end) { num1[2*i] = num[begin]; break; } num1[2*i] = num[end--]; num1[2*i+1] = num[begin++]; } for(int i=0; i; System.out.println...
    import java.util.Arrays;
    public class Sort {
    	public static void main(String[] args) {
    		int num[] = {99, 55, 77,33,44,22,11,66,88,0,100};
    		Arrays.sort(num);		
    		int[] num1 = new int[num.length];
    		int begin = 0;
    		int end = num.length - 1;
    		for(int i=0; i<(num.length+1)/2; i++)
    		{
    			//begin==end说明数组为奇数个
    			if(begin==end)
    			{
    				num1[2*i] = num[begin];
    				break;
    			}
    			num1[2*i] = num[end--];
    			num1[2*i+1] = num[begin++];
    		}	
    		for(int i=0; i<num1.length; System.out.println(num1[i++]));
    	}
    }


    展开全文
  • 以下这个2020网格四个处于同一对角线上相邻数字用红色标了出来: 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 ...

    在以下这个20×20的网格中,四个处于同一对角线上的相邻数字用红色标了出来:

    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

    这四个数字的乘积是:26 × 63 × 78 × 14 = 1788696.

    在这个20×20网格中,处于任何方向上(上,下,左,右或者对角线)的四个相邻数字的乘积的最大值是多少?

    public class Test {
    	//从左到右计算最大值
    	public static long leftToRight(int arr[][]){
    		int max = 0;
    		int value = 1;
    		for(int i = 0; i < 20; i ++){
    			for(int j = 0 ;j < 17 ; j++){
    				value = arr[i][j] * arr[i][j+1] * arr[i][j+2] * arr[i][j+3];
    				if(value > max) max = value;
    			}
    			
    		}
    		return max;
    	}
    	//从上到下计算最大值
    	public static long upToDown(int arr[][]){
    		int max = 0;
    		int value = 1;
    		for(int i = 0; i < 17; i ++){
    			for(int j = 0 ;j < 20 ; j++){
    				value = arr[i][j] * arr[i+1][j] * arr[i+2][j] * arr[i+3][j];
    				if(value > max) max = value;
    			}
    			
    		}
    		return max;
    		
    	}
    	//从左上到右下计算最大值
    	public static long left_uToRight_d(int arr[][]){
    		int max = 0;
    		int value = 1;
    		for(int i = 0; i < 17; i ++){
    			for(int j = 0;j < 17 ; j++){
    				value = arr[i][j] * arr[i+1][j+1] * arr[i+2][j+2] * arr[i+3][j+3];
    				if(value > max) max = value;
    			}
    			
    		}
    		return max;
    	}
    	//从右上到左下计算最大值
    	public static long right_uToLeft_d(int arr[][]){
    		int max = 0;
    		int value = 1;
    		for(int i = 3; i < 20; i ++){
    			for(int j = 0 ;j < 17 ; j++){
    				value = arr[i][j] * arr[i-1][j+1] * arr[i-2][j+2] * arr[i-3][j+3];
    				if(value > max) max = value;
    			}
    			
    		}
    		return max;
    	}
    	public static void main(String[] args) {
    		/* 构建20*20整数数组 */
    		int arr[][] = {{8,2,22,97,38,15,0,40,0,75,4,5,7,78,52,12,50,77,91,8},  
    					  {49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,4,56,62,0},  
    					  {81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,3,49,13,36,65},  
    					  {52,70,95,23,4,60,11,42,69,24,68,56,1,32,56,71,37,2,36,91},  
    					  {22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},  
    					  {24,47,32,60,99,3,45,2,44,75,33,53,78,36,84,20,35,17,12,50},  
    					  {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},  
    					  {24,55,58,5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},  
    					  {21,36,23,9,75,0,76,44,20,45,35,14,0,61,33,97,34,31,33,95},  
    				      {78,17,53,28,22,75,31,67,15,94,3,80,4,62,16,14,9,53,56,92},  
    					  {16,39,5,42,96,35,31,47,55,58,88,24,0,17,54,24,36,29,85,57},  
    					  {86,56,0,48,35,71,89,7,5,44,44,37,44,60,21,58,51,54,17,58},  
    					  {19,80,81,68,5,94,47,69,28,73,92,13,86,52,17,77,4,89,55,40},  
    					  {4,52,8,83,97,35,99,16,7,97,57,32,16,26,26,79,33,27,98,66},  
    					  {88,36,68,87,57,62,20,72,3,46,33,67,46,55,12,32,63,93,53,69},  
    					  {4,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36},  
    					  {20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,4,36,16},  
    					  {20,73,35,29,78,31,90,1,74,31,49,71,48,86,81,16,23,57,5,54},  
    					  {1,70,54,71,83,51,54,69,16,92,33,48,61,43,52,1,89,19,67,48},  
    	};  
    		System.out.println("max :" + max(leftToRight(arr),upToDown(arr),left_uToRight_d(arr),right_uToLeft_d(arr)));
    	}
    	private static long max(long a, long b,
    			long c, long d) {
    		long max = a;
    		if(b > max) max = b;
    		if(c > max) max = c;
    		if(d > max) max = d;
    		return max;
    	}
    	
    }
    


    展开全文
  • 以下这个2020网格四个处于同一对角线上相邻数字用红色标了出来: 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 0

    这是一个稍微有难度的题目,容易让人漏掉某种情况,我们先来看一下题目:

    在以下这个20×20的网格中,四个处于同一对角线上的相邻数字用红色标了出来:

    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

    这四个数字的乘积是:26 × 63 × 78 × 14 = 1788696.

    在这个20×20网格中,处于任何方向上(上,下,左,右或者对角线)的四个相邻数字的乘积的最大值是多少?

    这里最后一行的(上,下,左,右或者对角线)给你提供了思路,上下是一样的,左右也是一样的,但是对角线却又两条"左上--->右下“,”右上--->左下“。这两种情况是不一样的。我们通常都忘记了最后一种情况,导致最后做错,反正我刚开始就是这么做错的。

    好,我们来看一下程序:

    思路很简单,就是把20X20的放在一个二维数组中,然后分为四种情况分别进行运算就好了。

    use strict;
    use warnings;
    
    my $big=0;
    my $i;
    my $j;
    my $sum;
    my @num=(
    ["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"],
    );
    #第一种情况,左---->右
    for($i=0;$i<16;$i++)
    {
    	for($j=0;$j<16;$j++)
    	{
    		$sum = $num[$i][$j]*$num[$i][$j+1]*$num[$i][$j+2]*$num[$i][$j+3];
    		if($sum>$big)
    		{
    			$big=$sum;
    		}
    		else
    		{
    			next;
    		}
    	}
    }
    print "$big\n";
    
    
    #第二种情况,上--->下
    $big=0;
    for($i=0;$i<16;$i++)
    {
    	for($j=0;$j<16;$j++)
    	{
    		$sum = $num[$i][$j]*$num[$i+1][$j]*$num[$i+2][$j]*$num[$i+3][$j];
    		if($sum>$big)
    		{
    			$big=$sum;
    		}
    		else
    		{
    			next;
    		}
    	}
    }
    print "$big\n";
    
    #第三种情况,左上--->右下
    $big=0;
    for($i=0;$i<16;$i++)
    {
    	for($j=0;$j<16;$j++)
    	{
    		$sum = $num[$i][$j]*$num[$i+1][$j+1]*$num[$i+2][$j+2]*$num[$i+3][$j+3];
    		if($sum>$big)
    		{
    			$big=$sum;
    		}
    		else
    		{
    			next;
    		}
    	}
    }
    print "$big\n";
    
    #第四种情况,右上--->左下
    $big=0;
    for($i=4;$i<20;$i++)
    {
    	for($j=1;$j<16;$j++)
    	{
    		$sum = $num[$i][$j]*$num[$i-1][$j+1]*$num[$i-2][$j+2]*$num[$i-3][$j+3];
    		if($sum>$big)
    		{
    			$big=$sum;
    		}
    		else
    		{
    			next;
    		}
    	}
    }
    print "$big\n";
    
    


    得到四个结果,我们看看就知道大小了

    结果如下:

    C:\WINDOWS\system32\cmd.exe /c perl "F:\perl\c.pl"
    48477312
    51267216
    32719995
    70600674
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  • 要求按以下规则进行排序:第一个数最大,第二个数最小, 第三个数是剩下最大的,第四个数是剩下的最小的,依此类推 思路:每两个元素为一队进行比较,使之符合大的在前,小的在后的顺序,例如:.21 25 11 32 ...
  • 给定一个嵌套整数列表,返回列表所有整数总和,并按其深度加权。...因此,如果输入类似于[[1,1],2,[1,1]],则输出将为8,深度1为四个1,深度2为2。为了解决这个问题,我们将遵循以下步骤-定义一个功能dept...
  • 分治策略:是将规模比较大问题可分割成规模较小相同问题...分治法所能解决问题一般具有以下四个特征: 该问题规模缩小到一-定程度就可以容易地解决。 该问题可以分解为若干个规模较小相同问题。 使用小...
  • 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小...因为此时这两子数组各自的最大值为18,在所有情况最小。 动态规划 看到「
  • 4.限制了各个延时文本框的最大字7,显示列为4字节,以免字符串溢出错误。 5.★★★★添加了波特率可以随意手工设置为非标准值,例如 500bps,常用波特率仍然可选★★★★ 2005-3-8 1.修改发送程序,保证...
  • 410 分割数组的最大

    2020-07-25 09:49:03
    题目描述: 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小...因为此时这两子数组各自的最大值为18,在所有情况最小。 方法
  • 题目来源 题目描述 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小...因为此时这两子数组各自的最大值为18,在所有情况最小
  • 给定一非负整数组和一整数m,你需要将这数组分成m非空连续子数组。设计一算法使得这m个子数组各自和的最大值最小。...因为此时这两子数组各自的最大值为18,在所有情况最小。 ...
  • 以下是搜集关于SQL Server的最大容量规范,包含数据库引擎对象、实用工具对象、数据层引用对象、复制对象四个对象中的相关最大数量和最小数量。一般开发者重点还是在数据库引擎对象这块。个人觉得比较...
  • 题目 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小。...因为此时这两子数组各自的最大值为18,在所有情况最小。 解
  • 题目: 给定一非负整数组和一整数m,你需要将这数组分成m非空连续子数组。设计一算法使得这m个子数组各自和的最大值最小。...因为此时这两子数组各自的最大值为18,在所有情况最小...
  • 给定一非负整数组和一整数m,你需要将这数组分成m非空连续子数组。设计一算法使得这m个子数组各自和的最大值最小。...因为此时这两子数组各自的最大值为18,在所有情况最小。 分析...
  • 410. 分割数组的最大值 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值...因为此时这两子数组各自的最大值为18,在所有情况最小
  • 调查的是枝条的成熟度,每插条的节,每插条的叶子,芽的影响,生长素的浓度和生长素的溶剂。 枝条的成熟对成功有最大的影响,在活动末端生长的前30厘米以下采伐的插条可产生更多的根。 第二,使用增稠剂...
  • 题目: 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小。...因为此时这两子数组各自的最大值为18,在所有情况最小。 思路:
  • 给定一非负整数组和一整数m,你需要将这数组分成 m非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小。... 因为此时这两子数组各自的最大值为18,在所有情况
  • 分割数组的最大值 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小。...因为此时这两子数组各自的最大值为18,在所有情况.
  • 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小。...因为此时这两子数组各自的最大值为18,在所有情况最小。 来源:力
  • 概率统计——五概括法

    千次阅读 2017-08-18 14:51:54
    五数概括法 四分位数 ...五数概括法即用以下个数概括一组数据: 最小值 第一分位数 第二分位数(位数) 第三分位数 最大值 如下图所示: 其中我们将到位数距离大于1.5个IQR
  • 难度:困难 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小。...因为此时这两子数组各自的最大值为18,在所有情况最小。
  • **题干:** 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小。...因为此时这两子数组各自的最大值为18,在所有情况最小。.
  • 1. 题目 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小。...因为此时这两子数组各自的最大值为18,在所有情况最小。
  • 在MySQLBlob是一二进制对象,它是一可以存储大量数据容器(如图片,音乐等等),且能容纳不同大小数据,在MySQL有四种Blob类型,他们区别就是可以容纳信息量不容分别是以下四种: ①TinyBlob类型 ...
  • 题目 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小。...因为此时这两子数组各自的最大值为18,在所有情况最小。
  • 题干 给定一非负整数组和一整数 m,你需要将这数组分成 m 非空连续子数组。设计一算法使得这 m 个子数组各自和的最大值最小。...因为此时这两子数组各自的最大值为18,在所有情况最小。 想法
  • PTA黑洞(C语言版)

    2020-06-18 20:24:31
    输入在一行给出一三位。 输出格式: 按照以下格式输出重排求差过程: 序号: 数字重排后的最大数 - 重排后最小数 = 差值 序号从1开始,直到495出现在等号右边为止。 输入样例: 123 输出样例: 1: 321 -...

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