[C/C++] BOJ #1041 - Dice

This problem can be easily solved if you can visualize the geometric structure of a cube. Perhaps that’s why its difficulty level is quite low. The difficulty is Silver III. While it is harder than Bronze-level problems, it’s not too difficult.

As of now, the correct answer rate is 22.6%, which is quite low, and there are 531 correct submissions.

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The goal of the problem is to calculate the minimum possible sum of the visible faces when dice are arranged to form a cube of size \(N \times N \times N\), using standard 6-sided dice. Each edge of the cube consists of N dice.

I approached the problem by dividing it into two cases: when N = 1 and when \(N \geq 2\).
• For N = 1, since five faces of the die are visible, you can place the largest number on the bottom, which will be hidden.
• For \(N \geq 2\), you can organize the solution using a specific table of values.

 

sum of dices' faces

Once you derive this table, the problem becomes much more manageable. The reason I separated the N = 1 case is that the table only applies when \(N \geq 2\).

visible dice faces	number of dices
3	4
2	\(8N-12\)
1	\(5N^2 -16N + 12\)

 

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Here are the answers for several input values.

Input1:
1
1 2 3 4 5 6

15

Input2:
2
1 3 5 7 9 11

52

Input3:
987654
31 33 35 37 39 41

151196381478444

 

Note: Since the input numbers can be large, you should use the long long data type.

There aren’t many exceptional cases to handle.

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The following is the source code I wrote. Please use it for reference only.

//----------------------------------------------------------
//    baekjoon #1041 - Dice
//        - by Aubrey Choi
//        - created at 2019-12-06
//----------------------------------------------------------
#include <stdio.h>

int main()
{
    int n, v[6], i, j, sum=0, max=0, min=200, min2=200, min3=200;
    scanf("%d",&n);
    for(i=0;i<6;i++) scanf("%d",v+i);
    for(i=0;i<6;i++) sum+=v[i], max=v[i]>max?v[i]:max, min=v[i]<min?v[i]:min;
    if(n==1) { printf("%d\n", sum-max); return 0; }
    int p3[8][3] = { 0,1,2, 0,2,4, 0,3,4, 0,1,3, 1,2,5, 1,3,5, 2,4,5, 3,4,5 };
    int p2[12][2] = { 0,1, 0,2, 0,3, 0,4, 1,2, 1,3, 1,5, 2,4, 2,5, 3,4, 3,5, 4,5 };
    for(i=0;i<8;i++)
    {
        for(j=0,sum=0;j<3;j++) sum+=v[p3[i][j]];
        if(sum<min3) min3=sum;
    }
    for(i=0;i<12;i++)
    {
        for(j=0,sum=0;j<2;j++) sum+=v[p2[i][j]];
        if(sum<min2) min2=sum;
    }
    printf("%lld\n", 4*min3+(8LL*n-12)*min2+(5LL*n*n-16*n+12)*min);
}