盒子
盒子
文章目录
  1. 题目
  2. 题意
  3. 思路
  4. 代码
  5. 更新日志

Project Euler 11 Largest product in a grid(暴力)

题目

源地址:https://projecteuler.net/problem=11

题意

给定一个$20*20$的方格,求在同一条直线上(横、纵、对角)的四个数的最大乘积

思路

直接暴力找就行

代码

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#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
char str[] = {"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"};
int a[25][25];
int main () {
for (int i = 0; i < 20; i++) {
for (int j = 0; j < 20; j++) {
int k = i * 60 + j * 3;
a[i][j] = (str[k] - '0') * 10 + str[k + 1] - '0';
}
}
int mx = 0;
int dx[4] = {1, 1, 0, -1};
int dy[4] = {0, 1, 1, 1};
for (int i = 0; i < 20; i++) {
for (int j = 0; j < 20; j++) {
for (int k = 0; k < 4; k++) {
int ok = 1;
int x = i, y = j, now = a[x][y];
for (int l = 0; l < 3; l++) {
int tx = x + dx[k], ty = y + dy[k];
if (!(0 <= tx && tx < 20 && 0 <= ty && ty < 20)) {
ok = 0;
break;
}
now *= a[tx][ty];
x = tx, y = ty;
}
if (ok) mx = max(now, mx);
}
}
}
cout<<mx<<endl;
return 0;
}

更新日志

  • Completed on Mon, 27 Oct 2014, 12:13.31