D. XOR Determinant

You are given two arrays b and c of length n, consisting of non-negative integers. Construct n × n matrix
A as Aij = bi ⊕ cj . Find the determinant of A modulo 998 244 353

考虑 $A_{ij}=\sum_k b_{i,k}{c_{j,k}}+p$
其中 $c_{j,k}$ 为 $c_j$ 二进制第k位， $b_{i,k}$ 为 $b_i$ 二进制第k位为1时，为 $2^k$ ,否则为 $- 1$ ，并令 $p+=2^k$ 。
这样等价于矩阵每一位都可以用 $c_i$ 线性表示。
$log_2max{a_i}=60$
考虑第一列前62行，至少有一行可以写成其他61行的线性表示且系数与 $c_i$ 无关，所以该行其他列也可以用同列相同线性表示。也就是这一行可以由其它行线性表示得到。
因此矩阵 $r ank <= 61$

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (998244353)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 
ll pow2(ll a,int b,ll p)  //a^b mod p 
{  
    if (b==0) return 1%p;  
    if (b==1) return a%p;  
    ll c=pow2(a,b/2,p)%p;  
    c=c*c%p;  
    if (b&1) c=c*a%p;  
    return c%p;  
}  
ll inv(ll a,ll p) { //gcd(a,p)=1
	return pow2(a,p-2,p);
}
#define MAXN (100)
struct M  
{  
    int n,m;  
    ll a[MAXN][MAXN];  
    M(int _n=0){n=m=_n;MEM(a);}	
    M(int _n,int _m){n=_n,m=_m;MEM(a);}
    void mem (int _n=0){n=m=_n;MEM(a);}
    void mem (int _n,int _m){n=_n,m=_m;MEM(a);}
    ll mat[MAXN][MAXN],tmp[MAXN];
    ll det()
    {
    	For(i,n) For(j,m) mat[i][j]=a[i][j]%F; 
    	ll ans=1;
    	For(i,n)
    	{
    		int pos=i;
    		while (mat[pos][i]==0&&pos<n) ++pos;
    		if (mat[pos][i]==0) continue;
    		if (pos^i)
    		{
    			copy(mat[pos]+1,mat[pos]+1+m+1,tmp+1);
    			copy(mat[i]+1,mat[i]+1+m+1,mat[pos]+1);
    			copy(tmp+1,tmp+1+m+1,mat[i]+1);
    			ans=sub(0,ans);
    		}
    		ll invmatii=inv(mat[i][i],F);
			Fork(j,i+1,n)
				if (i^j)
				{
					ll p = mul(mat[j][i],invmatii);
					For(k,m) mat[j][k]=sub(mat[j][k],mul(mat[i][k],p));
				} 
    	}
    	For(i,n) ans=mul(ans,mat[i][i]);
    	return ans;
    }
}A;
#define eps 1e-6
struct M2  
{  
    int n,m;  
    ll a[MAXN][MAXN];  
    M2(int _n=0){n=m=_n;MEM(a);}	
    M2(int _n,int _m){n=_n,m=_m;MEM(a);}
    void mem (int _n=0){n=m=_n;MEM(a);}
    void mem (int _n,int _m){n=_n,m=_m;MEM(a);}
    
    long double mat[MAXN][MAXN],tmp[MAXN];
    long double det()
    {
    	For(i,n) For(j,m) mat[i][j]=a[i][j]; 
    	For(i,n)
    	{
    		int pos=i;
    		while (fabs(mat[pos][i])<eps&&pos<n) ++pos;
    		if (fabs(mat[pos][i])<eps) continue;
    		if (pos^i)
    		{
    			copy(mat[pos]+1,mat[pos]+1+m+1,tmp+1);
    			copy(mat[i]+1,mat[i]+1+m+1,mat[pos]+1);
    			copy(tmp+1,tmp+1+m+1,mat[i]+1);
    		}
			Fork(j,i+1,n)
				if (i^j)
				{
					long double p = mat[j][i]/mat[i][i];
					For(k,m) mat[j][k]-=mat[i][k]*p;
				} 
    	}
    	long double ans=1;
    	For(i,n) ans*=mat[i][i];
    	return ans;
    }
}A2;
int main()
{
//	freopen("D.in","r",stdin);
	int T;
	cin>>T;
	while(T--) {
		int n;
		cin>>n;
		if(n>=100) {
			For(i,2*n)read();
			puts("0");
		}
		else {
			ll b[100],c[100];
			For(i,n) cin>>b[i];
			For(i,n) cin>>c[i];
			A.mem(n);
			For(i,n) For(j,n) A.a[i][j]=(b[i]^c[j])%F; 
			cout<<A.det()<<endl;
			
		}	
	}
	
	
	
	
	return 0;
}

E. Egor Has a Problem

在这里插入图片描述
显然 $a_q/a_p\ge 1$ ，由于 $max{a_i}$ 不大，故存在上界 $c$ ，前 $c$ 个数里一定有至少2组相邻对 $a_q/a_p=1$

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 
int n;
ll a[500100];
void work() {
	For(i,n) Fork(j,i+1,n) Fork(k,j+1,n) Fork(l,k+1,n) {
		if(a[j]/a[i]==a[l]/a[k]){
			puts("YES");
			printf("%d %d %d %d\n",i,j,k,l);
			return;
		}
	}
	puts("NO");
}
void work2() {
	vector<int> v;
	For(i,n-1) {
		if(a[i+1]/a[i]==1) {
			v.pb(i);
		}
	}
	int sz=v.size();
	if(sz>=2 && v[0]+1<v[sz-1]){
		puts("YES");
		printf("%d %d %d %d\n",v[0],v[0]+1,v[sz-1],v[sz-1]+1);
	}
	else puts("NO");
}
int main()
{
//	freopen("E.in","r",stdin);
//	freopen(".out","w",stdout);
	n=read();
	For(i,n) cin>>a[i];
	if(n<=30) work();
	else work2();
	return 0;
}