2021 ICPC Asia Jinan Regional Contest-J Determinant (modulo Gaussian elimination)

topic link

https://pintia.cn/market/item/1459833348620926976

topic

meaning of the title

gives you an $n\times A\; matrix\; of\; n$ , and gives you a matrix oflength1$10^{}$delt deltwithin${}^4$$delt$ represents theabsolute value, and then we require the positive or negative of the value of the determinant of this matrix, if it is a positive number, output+otherwise output-

ideas

In fact, it is an ordinary Gaussian elimination, but there is one more modulo operation, because $The\; data\; of\; delt$ is very large, and if the floating point type is used, there will be overflow and precision problems, so we need to choose a modulus (prime number) by ourselves, here I choose$10^9+7$ Then we will read the$delt$ keeps taking the modulo, and then after running a modulo Gaussian elimination, we only need to judge whether the value of the determinant is the same as the$delt\; is$ the same, because what we read is an absolute value, if the value of the determinant we calculate is a negative number, because we have a modulo operation, it will definitely make its absolute value and negative number plus The value after the modulus is different, and for the operation of Gaussian elimination, you can refer to this:https://acmer.blog.csdn.net/article/details/122932692

code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define endl "\n"
#define PII pair<int,int>
#define INF 0x3f3f3f3f

const int N = 1e2+10;

ll n,a[N][N],delt;

ll ksm(ll a,ll b){
    
    
	ll ans = 1;
	while(b){
    
    
		if(b & 1) ans = ans * a % mod;
		b >>= 1;
		a = a * a % mod;
	}
	return ans;
}

ll inv(ll x){
    
    
	return ksm(x,mod-2);
}

void guss(){
    
    
	ll ans = 1LL;
	for(int i = 1;i <= n; ++i) {
    
    //枚举当前处理到第几列
		for(int j = i;j <= n; ++j) {
    
    //找到一个第i列不为空的行
			if(a[j][i]) {
    
    
				for(int k = i;k <= n; ++k) //交换i,j行
					swap(a[i][k],a[j][k]);
				if(i != j) ans = -ans;
				break;
			}
		}
		if(!a[i][i]) return ;
		//这里就是将第i行下面所有行的第i列清空，变成一个阶梯型的矩阵
		for(ll j = i + 1,iv = inv(a[i][i]);j <= n; ++j) {
    
    
			ll t = a[j][i] * iv % mod;
			for(ll k = i;k <= n; ++k) {
    
    
				a[j][k] = (a[j][k] - t * a[i][k] % mod + mod) % mod;
			}
		}
		//处理完第i行下面的所有行，于是我们将其a[i,i]乘在ans中
		ans = (ans * a[i][i] + mod) % mod;
	}
	//如果说我们通过取模计算的结果和delt相同，那么就说明矩阵的行列式是正的
	//因为如果是负数的话由于我们算的是行列式，而delt给的是行列式的绝对值，那么肯定会有不同
	if(ans == delt) cout<<"+"<<endl;
	else cout<<"-"<<endl;
}

int main()
{
    
    
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	string s;
	int t;
	cin>>t;
	while(t--){
    
    
		cin>>n>>s;
		delt = 0LL;
		for(int i = 0,l = s.size();i < l; ++i) {
    
    
			delt = (delt * 10LL + s[i] - '0') % mod;
		}
		for(int i = 1;i <= n; ++i) 
			for(int j = 1;j <= n; ++j)
				cin>>a[i][j];
		guss();
	}
	

	return 0;
}