A Linear Algebra Problem（唯一性的判定）
A Linear Algebra Problem
Time Limit: 3000/1000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)
God Kufeng is the God of Math. However, Kufeng is not so skilled with linear algebra, especially when dealing with matrixes.
One day, Captain Chen has a problem with matrix, here is the problem:
Given a n×n matrix A, what is the solution of n×n matrix X for the equation AX+XA=2A?
Captain Chen is a nice Captain, he wants to solve the equation only when A is a diagonal matrix, which means Aij=0 holds for all i≠j .
“That’s easy!” says Kufeng, “the answer is simply X=I, when I is the Identity Matrix.”
“But… is it the only solution for the equation above?” Captain Chen asks.
Kufeng cannot answer this question, can you help him?
Input
The first line of input is a number n, giving the size of matrix A and X. (1≤n≤1000)
Then comes a single line with n numbers, x1,x2,⋯,xn, where xi is the value of Aii. (−10000≤xi≤10000)
Output
If the answer is unique, output UNIQUE, otherwise output NOT UNIQUE
Sample input and output
| Sample Input | Sample Output |
|---|---|
3 1 2 3 |
UNIQUE |
2 1 -1 |
NOT UNIQUE |
Hint
For the second sample input, A=(100−1), there can be more than one possible solutions for X, for example, X=(1001) and X=(1101) both satisfy the equation, so the answer is not unique.
题目大意：
给你一个n维的对角矩阵A,然后有一个n维的矩阵X，使得AX+XA=2A的等式成立的条件下，要让X唯一，那么A应该满足什么样的条件？
解题思路：
刚读完题目后，以为是个找规律的题目，但是一想，还是错了，其实就是简单的矩阵推导了。
首先，对于单位阵X，那么一定有AX+XA=2A成立。现在来考虑，一般的情况。
令2A=B，那么，B[i,i] = A[i,i]*X[i,i]+X[i,i]*A[i,i] = 2*A[i,i]*X[i,i] = 2*A[i,i] ->如果要使X唯一，那么A[i,i]!=0.
B[i,j] = A[i,i]*X[i,j]+X[i,j]*A[j,j] = X[i,j]*(A[i,i]+A[j,j])=0 要使得X唯一，那么(A[i,i]+A[j,j])!=0.
这样一来，我们就发现了，在A的对角线上的元素不能有0和互为相反数的数了。
代码：
# include<cstdio>
# include<iostream>
# include<algorithm>
using namespace std;
# define MAX 1000+4
int a[MAX];
int main(void)
{
int n;cin>>n;
int flag = 0;
for ( int i = 0;i < n;i++ )
{
cin>>a[i];
if ( a[i]==0 )
{
flag = 1;
}
}
if ( !flag )
{
sort(a,a+n);
for ( int i = 0;i < n;i++ )
{
if ( a[i]>0||flag == 1 )
break;
for ( int j = n-1;j >= i;j-- )
{
if ( a[j]+a[i]==0 )
{
flag = 1;
break;
}
else if ( a[j] < 0 )
break;
else
continue;
}
}
}
if ( flag )
cout<<"NOT UNIQUE"<<endl;
else
cout<<"UNIQUE"<<endl;
return 0;
}