【高斯消元】2021ICPC济南补题-J Determinant

题目链接

Description

Alice uses the excellent property of the matrix $A^TA$ to find the determinant of the matrix $A^TA$. Recall that the determinant of $A$, denoted by $det(A)$, satisfies $det(A^TA)=det(A)^2$. Alice uses this property to find the absolute value $∣det(A)∣$. But unfortunately, when $∣det(A)≠0$, this method does not work out whether $det(A)$ is positive or negative.

Now that you know the matrix $A$ and the absolute value of $det(A)$, determine whether it is positive or negative.

Input

An integer $T (1≤T≤100)$ in the first line represents the number of test cases.

For each test case, the first line has an integer $n (1≤n≤100)$, where the size of matrix $A$ is $n×n$.

The second line has a large number $∣det(A)∣$. And it can be proved that $∣det(A)∣$ has no more than $10^4$ bits under the conditions in this problem.

The third to the $(n+2)$-th lines, each with $n$ numbers, describe this matrix. It is ensured that the absolute value of each number does not exceed $10^9$.

Output

For each test case, output a single line with a character "+" or "-", indicating whether the determinant is positive or negative.

Sample Input

3
1
1
1
2
2
1 2
3 4
2
5
-1 2
3 -1

Sample Output

+
-
-

题意

$T$组数据，给定一个$n*n$的矩阵和它的行列式的绝对值，要求判断其行列式的值正负，数据范围 $n (1≤n≤100)$ ,$T (1≤T≤100)$，$det(A)<=10^{10^4}$

思路

（当时说巧不巧刚好没听那周的数论班讲的高斯消元，赛中有三道数学题要么知识点不会要么不会推理，属实是又痛苦又后悔，补题也痛苦死了...***,退钱）

首先，逐字符用模数处理$det(A)$，变成一个小一点的可以比较的数，再计算行列式的值，对比一下是不是相等就知道是不是正数或者负数。

接着问题变成了怎么算行列式的值。又不能直接暴力解线性方程组。

这时候有一个很实用的数学方法，高斯消元，把矩阵消成一个阶梯的形状。

利用线性方程组实现过程：

一个$n*m$的矩阵，先找第一列是不是全为0，（如果是就跳过，跳到第二列）如果不是，找到第一个不为零的元素与第一行进行交换，（用线性代数里教的方法）用$a_{11}$把除其自身以外的该列所有元素消成0，其余同理，最终得到一个上三角行列式，从最后一行往上递推求解。

有一个很重要的点就是，必须取模，有些队伍带的板子没有取模的，据说这样也卡了一些人...
复杂度$O(n^3)$。

代码

#include <bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3f
#define int long long
#define ull unsigned long long
#define PII pair<int,int>
#define endl '\n'
const int N = 100 + 10;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);
typedef long long ll;
using namespace std;
int t, n;
int a[N][N];
string s;
int quickpow(int a, int b) {//快速幂
	int res = 1;
	while (b > 0) {
		if (b & 1) res = res * a % mod;
		b >>= 1;
		a = a * a % mod;
	}
	return res;
}
int getinv(int a) { return quickpow(a, mod - 2);}//费马小定理求逆元
int gauss(int n) {//高斯消元带模数
	int res = 1;
	for (int i = 1; i <= n; i++) {
		for (int j = i; j <= n; j++) {
			if (a[j][i]) {
				for (int k = i; k <= n; k++) {
					swap(a[i][k], a[j][k]);
				}
				if (i != j) {
					res = -res;
				}
				break;
			}
		}
		if (!a[i][i]) return 0;
		for (int j = i + 1, inv = getinv(a[i][i]); j <= n; j++) {
			int t = a[j][i] * inv % mod;
			for (int k = i; k <= n; k++) {
				a[j][k] = (a[j][k] - t * a[i][k] % mod + mod) % mod;
			}
		}
		res = (res * a[i][i] % mod + mod) % mod;
	}
	return res;
}
void init() {
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= n; j++) {
			a[i][j] = 0;
		}
	}
	return;
}
void solve() {
	cin >> n;
	cin >> s;//由于题面说了det_a是一个不超过1e4长度的大数，数值肯定存不下，所以这里要先读入字符串再用模数处理一下
	init();//初始化
	int det_a = s[0] - '0';
	for (int i = 1; i <= s.length() - 1; i++) {
		det_a = ((det_a * 10) % mod + (s[i] - '0')) % mod;
	}//处理大数
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cin >> a[i][j];
		}
	}//读入矩阵
	int sum = gauss(n);//高斯消元计算矩阵的值
	if (sum == det_a) cout << "+" << endl;//正数绝对值与自身相同，输出+
	else cout << "-" << endl;//负数绝对值与自身不同，输出-
	return;
}
signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	while (cin >> t) {
		while (t--) {
			solve();
		}
	}
	return 0;
}