Hdu 4920 Matrix multiplication (efficient)

Hdu 4920 Matrix multiplication (efficient)

Link: 4920 Matrix multiplication

Given two n-order matrices, evaluate the modulo after matrix multiplication 3.

Solution: there are only three situations after matrix modulo 3: 0, 1, and 2. Therefore, for matrix A, the location of 1 and 2 in each row is recorded with bitset. Position of 1, 2 in each column of matrix B. Then, for each position in the result, consider 1? 1, 1? 2, 2? 1, 2? 2.

#include 
  
   #include 
   
    #include 
    
     #include using namespace std;const int maxn = 805;int N, C[maxn][maxn];bitset
     
       x[maxn][2], y[maxn][2];void init () {    int u;    memset(C, 0, sizeof(C));    for (int i = 0; i < N; i++) {        for (int j = 0; j < 2; j++) {            x[i][j].reset();            y[i][j].reset();        }    }    for (int i = 0; i < N; i++) {        for (int j = 0; j < N; j++) {            scanf("%d", &u);            u %= 3;            if (u)                x[i][u-1].set(j, 1);        }    }    for (int i = 0; i < N; i++) {        for (int j = 0; j < N; j++) {            scanf("%d", &u);            u %= 3;            if (u)                y[j][u-1].set(i, 1);        }    }}int solve (int u, int v) {    int ret = 0;    for (int i = 0; i < 2; i++) {        for (int j = 0; j < 2; j++) {            bitset
      
        k = x[u][i]&y[v][j]; ret = (ret + (i+1)*(j+1)*k.count()) % 3; } } return ret;}int main () { while (scanf("%d", &N) == 1) { init(); for (int i = 0; i < N; i++) { printf("%d", solve(i, 0)); for (int j = 1; j < N; j++) printf(" %d", solve(i, j)); printf("\n"); } } return 0;}