共计 1895 个字符,预计需要花费 5 分钟才能阅读完成。
problem11
地址:https://projecteuler.net/problem=11。
源码:git@code.aliyun.com:c-program/projecteuler.git。
问题:找出 20X20 表格中上下左右乘积最大值。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include "debug.h"
#define NUM 20
int main(int argc, char **argv){
char *strsrc= "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";
char *sep = " ";
char *s = NULL;
int list[NUM][NUM];
int i, j;
char *tmp;
long int iResult = 0;
long int tmpResult = 0;
debugTime();
s = strdup(strsrc);
tmp = strsep(&s, sep);
i = 0;
while (NULL != tmp){list[i / NUM][i % NUM] = atoi(tmp);
tmp = strsep(&s, sep);
i++;
}
for (i = 0; i < NUM; i++){for (j = 0; j < NUM; j++){if (0 == list[i][j]) continue;
if (NUM > j + 3){tmpResult = list[i][j] * list[i][j + 1] * list[i][j + 2] * list[i][j + 3];
iResult = iResult>tmpResult?iResult:tmpResult;
}
if (NUM > i + 3){tmpResult = list[i][j] * list[i + 1][j] * list[i + 2][j] * list[i + 3][j];
iResult = iResult>tmpResult?iResult:tmpResult;
}
if ((NUM > i + 3) && (NUM > j + 3)){tmpResult = list[i][j] * list[i + 1][j + 1] * list[i + 2][j + 2] * list[i + 3][j + 3];
iResult = iResult>tmpResult?iResult:tmpResult;
}
if ((NUM > i + 3) && (NUM > j - 3)){tmpResult = list[i][j] * list[i + 1][j - 1] * list[i + 2][j - 2] * list[i + 3][j - 3];
iResult = iResult>tmpResult?iResult:tmpResult;
}
}
}
printf("Problem11 Answer: %ld\n",iResult);
debugTime();
return 0;
}
正文完