Codeforces Round #230 (Div. 2)B. Three matrices

B. Three matrices

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Chubby Yang is studying linear equations right now. He came up with a nice problem. In the problem you are given an n × n matrix W, consisting of integers, and you should find two n × n matrices A and B, all the following conditions must hold:

• Aij = Aji, for all i, j (1 ≤ i, j ≤ n);
• Bij =  - Bji, for all i, j (1 ≤ i, j ≤ n);
• Wij = Aij + Bij, for all i, j (1 ≤ i, j ≤ n).

Can you solve the problem?

Input

The first line contains an integer n (1 ≤ n ≤ 170). Each of the following n lines contains n integers. The j-th integer in the i-th line is Wij(0 ≤ |Wij| < 1717).

Output

The first n lines must contain matrix A. The next n lines must contain matrix B. Print the matrices in the format equal to format of matrix Win input. It is guaranteed that the answer exists. If there are multiple answers, you are allowed to print any of them.

The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 4.

Sample test(s)

input

21 43 2

output

1.00000000 3.500000003.50000000 2.000000000.00000000 0.50000000-0.50000000 0.00000000

input

31 2 34 5 67 8 9

output

1.00000000 3.00000000 5.000000003.00000000 5.00000000 7.000000005.00000000 7.00000000 9.000000000.00000000 -1.00000000 -2.000000001.00000000 0.00000000 -1.000000002.00000000 1.00000000 0.00000000

解题报告：

此题为纯粹数学题。代码如下：

#include <iostream>#include <cstdio>using namespace std;double w[175][175],a[175][175],b[175][175];int main(){    int n;    scanf("%d",&n);    for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)            scanf("%lf",&w[i][j]);    for(int i=1;i<=n;i++)        for(int j=1;j<=i;j++){            a[i][j]=(w[i][j]+w[j][i])/2;            a[j][i]=a[i][j];            b[i][j]=w[i][j]-a[i][j];            b[j][i]=b[i][j]-2*b[i][j];        }    for(int i=1;i<=n;i++){        for(int j=1;j<=n;j++){            if(j==1)                printf("%.8lf",a[i][j]);            else                printf(" %.8lf",a[i][j]);        }        printf("\n");    }    for(int i=1;i<=n;i++){        for(int j=1;j<=n;j++){            if(j==1)                printf("%.8lf",b[i][j]);            else                printf(" %.8lf",b[i][j]);        }        printf("\n");    }    return 0;}