当前位置: 代码迷 >> J2SE >> 欧拉计划上的一道算法题。求解,该怎么处理
  详细解决方案

欧拉计划上的一道算法题。求解,该怎么处理

热度:101   发布时间:2016-04-24 01:01:08.0
欧拉计划上的一道算法题。。求解
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?


------解决方案--------------------
自己遍历计算一下就好了

for example

Java code
import java.util.*;public class Test {    public static void main(String[] args) throws Throwable {        int[][] matrix = {            { 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, 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,  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}        };        Map<Integer, int[]> map = getMaxM(matrix, 4);        for (Map.Entry<Integer, int[]> e : map.entrySet()) {            for (int n : e.getValue()) {                System.out.printf("%02d ", n);            }            System.out.println(e.getKey());        }    }    public static Map<Integer, int[]> getMaxM(int[][] matrix, int size) {        Map<Integer, int[]> map = new LinkedHashMap<Integer, int[]>();         int[] tmp = new int[size];        int m = 0, max = Integer.MIN_VALUE;        for (int i=0; i<matrix.length; i++) {            for (int j=0; j<matrix[i].length; j++) {                if (j<=matrix[i].length-size) {                    m = 1;                    for (int k=0; k<size; k++) {                        tmp[k] = matrix[i][j+k];                        m *= tmp[k];                    }                    if (m > max) {                        max = m;                        map.clear();                        map.put(m, Arrays.copyOf(tmp, size));                    }                }                if (i<=matrix.length-size) {                    m = 1;                    for (int k=0; k<size; k++) {                        tmp[k] = matrix[i+k][j];                        m *= tmp[k];                    }                    if (m > max) {                        max = m;                        map.clear();                        map.put(m, Arrays.copyOf(tmp, size));                    }                }                                if (i<=matrix.length-size && j<=matrix[i].length-size) {                    m = 1;                    for (int k=0; k<size; k++) {                        tmp[k] = matrix[i+k][j+k];                        m *= tmp[k];                    }                    if (m > max) {                        max = m;                        map.clear();                        map.put(m, Arrays.copyOf(tmp, size));                    }                }                if (i<=matrix.length-size && j>=size-1) {                    m = 1;                    for (int k=0; k<size; k++) {                        tmp[k] = matrix[i+k][j-k];                        m *= tmp[k];                    }                    if (m > max) {                        max = m;                        map.clear();                        map.put(m, Arrays.copyOf(tmp, size));                    }                }             }        }        return map;    }}
  相关解决方案