Problem F: Failed in Linear Algebra ## Problem F: Failed in Linear Algebra

Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 587 Solved: 86

[Submit][Status][Web Board]## Description

One day, Wavator is taking his Linear algebra course. He hates calculating the expression of matrix so he wants to develop a calculator to help him. But, he got 59 in last years’ DSAA course, so he turns you for help.

n square matrices of size m are given, and we defines an operation like “(1+2)*1” which means the matrix 1 plus matrix 2 and then multiplies matrix 1.

Wavator only wants to calculate “+” and “-” and “*”, so he denotes that “+” means A + B = C where C[i][j] = A[i][j] + B[i][j] .The rule of “-” is similar with "+".

Notice that in matrix multiplication, a*b and b*a is not the same.

Since the number may be too large during the calculation process, in each step you should mod 1000000007.

## Input

The ﬁrst line contains an integer T, meaning there will be T (T<=10) test cases.

For each test cases, the ﬁrst line is n and m (n <= 10 and m <= 50), then there will be n parts, each part is a m * m matrix a, 0<=a[i][j] <= 10000, matrixes are numbered from 1 to n, then there will be a string s, the length of s is not greater than 50, and it is valid. (contains only 1~n numbers (index of matrixes) and “+”, “-”, “*” only) .

## Output

For each test case, print m lines, each line should contain m integers, meaning the value of the ﬁnal matrix at this line and this column.

## Sample Input

1
2 2
1 2
2 1
2 2
3 3
(1+2)*1

## Sample Output

11 10
13 14

## HINT

Codes of mod in c++ lang. similar using java.

const int MOD = 1000000007;

inline int add(int x, int y) { return (x + y) % MOD; }

inline int sub(int x, int y) { return (x - y + MOD) % MOD; }

inline int mul(int x, int y) { return static_cast<int>((long long) x * y % MOD); }

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