12ASuper Agent

3阶矩阵、判断中心对称

英语是硬伤，就是让我们判断对称矩阵是否相等，因为是三阶矩阵所以我们就直接判断。

                A. Super Agent                time limit per test2 seconds                memory limit per test256 megabytes                inputstandard input                outputstandard output

There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.

Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.

Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.

Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».

Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.

Examples
input
XX.
…
.XX
output
YES
input
X.X
X..
…
output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry

#include<cstdio>#include<iostream>#include<cstring>using namespace std;int main(){    char a[5][5];    for(int i=0; i<3; i++)        for(int j=0; j<3; j++)            cin>>a[i][j];    if(a[0][0]==a[2][2]&&a[0][1]==a[2][1]&&a[0][2]==a[2][0]&&a[1][0]==a[1][2])        puts("YES");    else        puts("NO");    return 0;}