Project Euler -> prob 4000 lem 11

11. 在以下这个2020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2020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？

int main(void)
{
int arr[20][20] = {
{ 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,  2, 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}
};
int i,num[5]={1,1,1,1,1};
for(i=0;i<20;i++){
int count=1,j=0;
num[4]=1;
while(j<=20){
if(count%5==0){
num[0]=num[0]>num[4]?num[0]:num[4];
num[4] = 1;
if(j<20)
j-=3;
else
break;
count++;
}
else{
num[4] *= arr[i][j];
j++;
count++;
}
}
}
for(i=0;i<20;i++){
int count=1,j=0;num[4]=1;
while(j<=20){
if(count%5==0){
num[1]=num[1]>num[4]?num[1]:num[4];
num[4] = 1;
if(j<20)
j-=3;
else
break;
count++;
}
else{
num[4] *= arr[j][i];
j++;
count++;
}
}
}
for(i=0;i<16;i++){
int count=1,j=i,k=0;num[4]=1;
while(k<=20){
if(count%5==0){
num[2] = num[2]>num[4]?num[2]:num[4];
num[4] = 1;
if(k<20){
j=i;
k-=3;
}
else
break;
count++;
}
else{
num[4]*= arr[j++][k++];
count++;
}
}
}
for(i=0;i<16;i++){
int count=1,j=i,k=19;num[4]=1;
while(k>=0){
if(count%5==0){
num[3] = num[3]>num[4]?num[3]:num[4];
num[4] = 1;
if(k>0){
j=i;
k+=3;
}
else
break;
count++;
}
else{
num[4]*= arr[j++][k--];
count++;
}
}
}
num[0]=num[0]>num[1]?num[0]:num[1];
num[0]=num[0]>num[2]?num[0]:num[2];
num[0]=num[0]>num[3]?num[0]:num[3];
printf("The maximum product is %d",num[0]);
return 0;
}

Answer:

70600674