Matrix operation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 907 Accepted Submission(s): 384
Problem Description
M_0 is interested in cryptography. Recently he learned the Advanced Encryption Standard (AES). AES operates on a 4×4 column-major order matrix of bytes and he found it that there is a step called the MixColumns step in the AES.
In the MixColumns step, there is a state matrix, the four bytes of each column of the state are combined using an invertible linear transformation. The MixColumns function takes four bytes as input and outputs four bytes, where each input byte affects all four output bytes. Together with ShiftRows, MixColumns provides diffusion in the cipher.
During this operation, each column is multiplied by the known matrix that for the 128 bit key is:
The addition operation is defined as: xor.
The multiplication operation is defined as:
multiplication by 1 means no change
multiplication by 2 means shifting to the left
multiplication by 3 means shifting to the left and then performing xor with the initial unshifted value.
Notice:After each shifting, a conditional xor with 0x1B should be performed if the shifted value is larger than 0xFF.
In the MixColumns step, there is a state matrix, the four bytes of each column of the state are combined using an invertible linear transformation. The MixColumns function takes four bytes as input and outputs four bytes, where each input byte affects all four output bytes. Together with ShiftRows, MixColumns provides diffusion in the cipher.
During this operation, each column is multiplied by the known matrix that for the 128 bit key is:
The addition operation is defined as: xor.
The multiplication operation is defined as:
multiplication by 1 means no change
multiplication by 2 means shifting to the left
multiplication by 3 means shifting to the left and then performing xor with the initial unshifted value.
Notice:After each shifting, a conditional xor with 0x1B should be performed if the shifted value is larger than 0xFF.
Input
There are several cases.
The first line is an integer T (T <= 20000), indicating the test cases.
Then is the state matrix, each case followed by four lines, each line contains four bytes, separated by spaces.
The first line is an integer T (T <= 20000), indicating the test cases.
Then is the state matrix, each case followed by four lines, each line contains four bytes, separated by spaces.
Output
For each case, output the new matrix of the state matrix after the MixColumns step. The output data for two different test cases should be separated by an empty line.
Sample Input
1
00 01 02 03
04 05 06 07
08 09 0A 0B
0C 0D 0E 0F
Sample Output
08 09 0A 0B
1C 1D 1E 1F
00 01 02 03
14 15 16 17
Author
FZU
Source
题目大意：两个矩阵“相乘”，输出相乘后的矩阵。这里给了一个key矩阵，表示秘钥矩阵。下面给出t组数据，每组都是4*4的矩阵，用key矩阵乘以matrix得到结果，矩阵不改变，只是用矩阵中的值进行操作，这里加法用Xor代替，如果key矩阵中是1，那么就不对matrix中的值操作，如果是2，就让matrix中的值左移一位，如果是3，就让matrix中的值左移一位，并且与没有左移的值进行异或。左移可能会超出8位能表示的最大值，所以如果结果大于0xFF，就跟0x1B异或，然后对256取余。结果是大写字母，所以输出时应该是%02X输出，不能用%02x输出。
#include<bits/stdc++.h>
using namespace std;
int key[4][4]={2,3,1,1,1,2,3,1,1,1,2,3,3,1,1,2};
int ini[4][4],ans[4][4];
int main(){
int t;
scanf("%d",&t);
while(t--){
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
scanf("%x",&ini[i][j]);
}
}
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
int a,tmp=0;
for(int k=0;k<4;k++){
if(key[i][k]==1){
a=ini[k][j];
}else if(key[i][k]==2){
a=ini[k][j];
a<<=1;
if(a>0xFF){
a^=0x1B;
a%=(0xFF+1);
}
}else if(key[i][k]==3){
a=ini[k][j];
a<<=1;
if(a>0xFF){
a^=0x1B;
a%=(0xFF+1);
}
a^=ini[k][j];
}
tmp^=a;
}
ans[i][j]=tmp;
}
}
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
printf("%02x%c",ans[i][j],j==3?'\n':' ');
}
}
if(t)
puts("");
}
return 0;
}