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ural 1433. Diamonds
小程序：扫一扫查出行【扫一扫了解最新限行尾号】
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1433. Diamonds
Time limit: 1.0 second
Memory limit: 64 MB
Sasha is lucky to have a diamond in the form of a regular tetrahedron. Incidentally, his friend Pasha has got a diamond of exactly the same shape and size. In order to distinguish between their diamonds, the friends decided to paint the crystals. Each of them painted each face of his diamond a certain color. The diamonds became very beautiful, so Pasha and Sasha were happy. But their happiness did not last long. That night, Sasha woke up with a sudden thought — what if his and Pasha's diamonds are still indistinguishable? He decided to call Pahsa immediately. Sasha ran up to the phone, tried to grasp the receiver, but at that moment the phone rang. Of course, it was Pasha, who had the same sudden thought. So Sasha and Pasha hastened to tell each other the colors of their diamonds' faces… Their worst fears were confirmed. Their diamonds were identical, and to see it one simply had to turn one of the diamonds.
You are to write a program that could prevent this horrible mistake. Given a scheme of the supposed coloring of the diamonds, determine if these colorings are identical, i.e., if one of them can be obtained from the other by turning the crystal.
Input
The input contains two lines. Each line contains four letters, which denote the colors of the faces in the following order: the base face, the "left front" face, the "right front" face, and the back face. There are only four paints available: red, green, blue, and yellow, denoted by the letters R, G, B, and Y, respectively.
Output
Output the word "equal" if the colored tetrahedrons will be identical, and the word "different" otherwise.
Sample
input
output
RGRB
GRRB
equal
Problem Author:
Pavel Egorov, Stanislav Vasilyev
Problem Source:
The 7th USU Open Personal Contest - February 25, 2006
Memory limit: 64 MB
Sasha is lucky to have a diamond in the form of a regular tetrahedron. Incidentally, his friend Pasha has got a diamond of exactly the same shape and size. In order to distinguish between their diamonds, the friends decided to paint the crystals. Each of them painted each face of his diamond a certain color. The diamonds became very beautiful, so Pasha and Sasha were happy. But their happiness did not last long. That night, Sasha woke up with a sudden thought — what if his and Pasha's diamonds are still indistinguishable? He decided to call Pahsa immediately. Sasha ran up to the phone, tried to grasp the receiver, but at that moment the phone rang. Of course, it was Pasha, who had the same sudden thought. So Sasha and Pasha hastened to tell each other the colors of their diamonds' faces… Their worst fears were confirmed. Their diamonds were identical, and to see it one simply had to turn one of the diamonds.
You are to write a program that could prevent this horrible mistake. Given a scheme of the supposed coloring of the diamonds, determine if these colorings are identical, i.e., if one of them can be obtained from the other by turning the crystal.
Input
The input contains two lines. Each line contains four letters, which denote the colors of the faces in the following order: the base face, the "left front" face, the "right front" face, and the back face. There are only four paints available: red, green, blue, and yellow, denoted by the letters R, G, B, and Y, respectively.
Output
Output the word "equal" if the colored tetrahedrons will be identical, and the word "different" otherwise.
Sample
| input | output |
|---|---|
RGRB |
equal |
Problem Author:
Pavel Egorov, Stanislav Vasilyev
Problem Source: The 7th USU Open Personal Contest - February 25, 2006
Problem Source: The 7th USU Open Personal Contest - February 25, 2006
Tags:
none (hide tags for unsolved problems)
Difficulty: 252
题意：给两个正四面的的四个面的颜色，判断两个四面体是否相等。
分析：所以我们只要将所有匹对方式打个表就好。
/**
Create By yzx - stupidboy
*/
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <ctime>
#include <iomanip>
using namespace std;
typedef long long LL;
typedef double DB;
#define For(i, s, t) for(int i = (s); i <= (t); i++)
#define Ford(i, s, t) for(int i = (s); i >= (t); i--)
#define Rep(i, t) for(int i = (0); i < (t); i++)
#define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
#define rep(i, x, t) for(int i = (x); i < (t); i++)
#define MIT (2147483647)
#define INF (1000000001)
#define MLL (1000000000000000001LL)
#define sz(x) ((int) (x).size())
#define clr(x, y) memset(x, y, sizeof(x))
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define ft first
#define sd second
#define mk make_pair
inline void SetIO(string Name)
{
string Input = Name+".in",
Output = Name+".out";
freopen(Input.c_str(), "r", stdin),
freopen(Output.c_str(), "w", stdout);
} inline int Getint()
{
int Ret = ;
char Ch = ' ';
bool Flag = ;
while(!(Ch >= '' && Ch <= ''))
{
if(Ch == '-') Flag ^= ;
Ch = getchar();
}
while(Ch >= '' && Ch <= '')
{
Ret = Ret * + Ch - '';
Ch = getchar();
}
return Flag ? -Ret : Ret;
} const int N = , M = ;
int Face[][]={
{, , , }, {, , , }, {, , , },
{, , , }, {, , , }, {, , , },
{, , , }, {, , , }, {, , , },
{, , , }, {, , , }, {, , , }
};
struct Node
{
int Arr[];
inline void Read()
{
string Str;
cin >> Str;
Rep(i, ) Arr[i] = Str[i];
}
} A, B; inline void Input()
{
A.Read();
B.Read();
} inline void Solve()
{
Rep(i, )
{
bool Flag = ;
Rep(j, )
if(A.Arr[Face[i][j]] != B.Arr[j])
{
Flag = ;
break;
}
if(!Flag)
{
puts("equal");
return;
}
}
puts("different");
} int main()
{
#ifndef ONLINE_JUDGE
SetIO("B");
#endif
Input();
Solve();
return ;
}
