Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

I  — Dice

Time Limit: 1 sec
Memory Limit: 32 MB

Mary add Sue are playing with dices. Rules are simple: at the begging each of them puts coin on the table and roll a dice. Wins a player who rolled a larger number. If numbers are the same, coins stay on the table for a next round. In order to make this game more interesting they decided to play now with normal dices, but with dice that can have arbitrary number of bones, from 0 till 9. However each round must be played with same dice by both players.

Girls have been playing this game for a day long, till Mary run out of coins (nevertheless she had more coins at the beginning of the game). Now Mary is confused. How could she have lost all her coins? She thinks that Sue had been cheating. Before each roll Mary wrote on a paper numbers of bones on each side of the dice. Now she wonders if same dice was always used during one round. Help her to find it out.

INPUT

On the first line there is the total number of test cases T (T ≤ 10³), next T lines follows. Each line contains two six digit numbers, each digit stands for number of bones on side of a dice in this order: top, bottom, front, left, back, right.

OUTPUT

For each test case output line "Equal" if two dices are equals, or "Not Equal" otherwise.

SAMPLE INPUT

3
345678 345678
123123 123456
123456 351624

SAMPLE OUTPUT

Equal
Not Equal
Equal

---

Problem by: Aleksej Viktorchik; Leonid Sislo
Huge Easy Contest #2

把翻法都依序找出來，然後找一個最小表示法去做比較即可。

#include <stdio.h>
#include <stdlib.h>
typedef struct {
    int dice[6];
} DICE;
void Right_turn(DICE *A) {
    static int t;
    t = A->dice[0], A->dice[0] = A->dice[4];
    A->dice[4] = A->dice[3], A->dice[3] = A->dice[1], A->dice[1] = t;
}
void Up_turn(DICE *A) {
    static int t;
    t = A->dice[0], A->dice[0] = A->dice[5];
    A->dice[5] = A->dice[3], A->dice[3] = A->dice[2], A->dice[2] = t;
}
void clockwise(DICE *A) {
    static int t;
    t = A->dice[2], A->dice[2] = A->dice[4];
    A->dice[4] = A->dice[5], A->dice[5] = A->dice[1], A->dice[1] = t;
}
int cmp(const void *a, const void *b) {
    static DICE *i, *j;
    static int k;
    i = (DICE *)a, j = (DICE *)b;
    for(k = 0; k < 6; k++)
        if(i->dice[k] != j->dice[k])
            return i->dice[k] - j->dice[k];
    return 0;
}

DICE Spin_dice(DICE A) {
    static int i, j;
    DICE B = A;
    for(i = 0; i < 4; i++) {
        for(j = 0; j < 4; j++) {
            if(cmp(&B, &A) > 0)
                B = A;
            clockwise(&A);
        }
        Right_turn(&A);
    }
    Up_turn(&A);
    for(i = 0; i < 2; i++) {
        for(j = 0; j < 4; j++) {
            if(cmp(&B, &A) > 0)
                B = A;
            clockwise(&A);
        }
        Up_turn(&A), Up_turn(&A);
    }
    return B;
}
main() {
    int n, i;
    DICE A, B;// 前右上後左下
    char a[20], b[20]; // 上下前左後右
    scanf("%d", &n);
    while(n--) {
        scanf("%s %s", a, b);
        A.dice[0] = a[2];
        A.dice[1] = a[5];
        A.dice[2] = a[0];
        A.dice[3] = a[4];
        A.dice[4] = a[3];
        A.dice[5] = a[1];

        B.dice[0] = b[2];
        B.dice[1] = b[5];
        B.dice[2] = b[0];
        B.dice[3] = b[4];
        B.dice[4] = b[3];
        B.dice[5] = b[1];
        A = Spin_dice(A);
        B = Spin_dice(B);
        for(i = 0; i < 6; i++)
            if(A.dice[i] != B.dice[i])
                break;
        if(i == 6)
            puts("Equal");
        else
            puts("Not Equal");
    }
    return 0;
}