现在的位置: 首页 > 综合 > 正文

Largest product in a grid( Project Euler problem 11)

2013年02月26日 ⁄ 综合 ⁄ 共 1881字 ⁄ 字号 评论关闭

In the 20×20 grid below, four numbers along a diagonal line have been marked in red.

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

The product of these numbers is 26 × 63 × 78 × 14
= 1788696.

What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?

http://projecteuler.net/problem=11

需要把矩阵分区,第0~19&&0~15列的求横向的,第0~15&&0~19列求纵向的,第0~15&&0~15列求斜向右下的,第0~15&&4~19列求斜向左下的。

#include<cstdio>
#include<iostream>
using namespace std;
int n[20][20];
int main()
{
    freopen("1.txt","r",stdin);
    for(int i=0; i<20; i++)
        for(int j=0; j<20; j++)
            cin>>n[i][j];
    int max=1;
    for(int i=0; i<20; i++)
        for(int j=0; j<16; j++)
            if (n[i][j]*n[i][j+1]*n[i][j+2]*n[i][j+3]>max )
                max=n[i][j]*n[i][j+1]*n[i][j+2]*n[i][j+3];
    for(int i=0; i<16; i++)
        for(int j=0; j<20; j++)
            if(n[i][j]*n[i+1][j]*n[i+2][j]*n[i+3][j]>max)
                max=n[i][j]*n[i+1][j]*n[i+2][j]*n[i+3][j];
    for(int i=0; i<16; i++)
        for(int j=0; j<16; j++)
            if(n[i][j]*n[i+1][j+1]*n[i+2][j+2]*n[i+3][j+3]>max)
                max=n[i][j]*n[i+1][j+1]*n[i+2][j+2]*n[i+3][j+3];
    for(int i=0; i<16; i++)
        for(int j=4; j<20; j++)
            if (n[i][j]*n[i+1][j-1]*n[i+2][j-2]*n[i+3][j-3]>max)
                max=n[i][j]*n[i+1][j-1]*n[i+2][j-2]*n[i+3][j-3];
    cout<<max<<endl;
    return 0;
}

  

抱歉!评论已关闭.