ICPC2020济南A Matrix Equation

写在前面

大概是高中退役之后第一次写写题解，当然要从基础写起，因为变得更菜了，所以只能写写更水的题目了

题目链接

ICPC济南A

题解

\(A \times B = C⊙B\)

\[B_{ij} \times C_{ij}=\sum_{k = 1}^{n}A_{ik} B_{kj} (mod 2) \]

对于每个\(j\)有

\[[i = 1,2,...n]\sum_{k = 1}^n (A_{ik} [k = i]xorC_{i})B_i \]

对于每个\(j\)都有\(n * n\)的异或方程组，求自由元设为\(k_j\)得到方程组的解的个数为\(2^{k_{j}}\)
每个\(j\)独立于是有\(ans = \prod_{j=1}^n2^{k_{j}}\)
复杂度\(n^4/32\)

代码

#include<cmath> 
#include<cstdio> 
#include<iostream> 
#include<algorithm> 
#include<bits/stdc++.h> 
#define rep(a,b,c) for(int a = b;a <= c;++ a) 
using namespace std; 
#define gc getchar() 
#define pc putchar
inline int read() { 
    int x = 0; 
    char c = gc; 
    while(c < '0' || c > '9') c = gc; 
    while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc; 
    return x; 
} 
const int maxn = 207; 
#define LL long long
const int mod = 998244353; 
int n; 
int A[207][207]; 
int B[207][207]; 
int type[maxn]; 

LL qpow(int k) {   
	if(k <= 0) return 1; 
     LL ret = 1,tmp = 2; 
    for(;k;k >>= 1) {  
        if(k & 1) ret *= tmp,ret %= mod; 
        tmp *= tmp,tmp %= mod; 
    }
    return ret; 
} 
bitset<330>a[305]; 
LL guass(int equ,int var) { 
    int r,c,t; 
    for(r = c = 1;r <= equ && c <= var;++ r,++ c) { 
        for(t = r;t <= equ;++ t) if(a[t][c]) break; 
        if(t == equ + 1) {
            -- r;continue; 
        } else swap(a[t],a[r]); 
        if(a[r][c] == 0) { r -- ;continue; }  
        for(int i = r + 1;i <= equ;++ i) if(a[i][c]) a[i] ^= a[r];  
        //for(int i = 1;i <= n;++ i) for(int k = 1;k <= n;++ k)  cout<<a[i][k];puts("") ; 
    } 
    return qpow(var + 1 - r); 
} 

int main() {
    n = read(); 
    rep(i,1,n) rep(j,1,n) A[i][j] = read(); 
    rep(i,1,n) rep(j,1,n) B[i][j] = read(); 
    LL ans = 1; 
    
	for(int j = 1;j <= n;++ j) { 
        rep(i,1,n) rep(k,1,n) a[i][k] = A[i][k]; 
		rep(i,1,n) a[i][i] = (A[i][i] ^ B[i][j]);
		 
        //rep(i,1,n) rep(k,1,n) cout<<a[i][k];puts("") ; 
    	ans *= guass(n,n); ans %= mod;  
    }cout<< ans; 
} 

关于抄之前代码发现之前代码写错了竟调了一中午这件事。。。