HDU 4920 Matrix multiplication（bitset)

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HDU 4920 Matrix multiplication

题目链接

题意：给定两个矩阵，求这两个矩阵相乘mod 3

思路：没什么好的想法，就把0的位置不考虑，结果就过了。然后看了官方题解，上面是用了bitset这个东西，可以用来存大的二进制数，那么对于行列相乘，其实就几种情况，遇到0都是0了，1 1得1，2 1，1 2得2，2 2得1，所以只要存下行列1和2存不存在分别表示的二进制数，然后取且bitcount一下的个数，就可以计算出相应的数值了

代码：

暴力：

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;inline void scanf_(int &num)//无负数{    char in;    while((in = getchar()) > '9' || in < '0') ;    num = in - '0';    while(in = getchar(),in >= '0' && in <= '9')num *= 10,num += in - '0';}const int N = 805;int n;int a[N][N], av[N][N], an[N], b[N][N], bv[N][N], bn[N], c[N][N];int main() {    while (~scanf("%d", &n)) {int num;memset(an, 0, sizeof(an));memset(bn, 0, sizeof(bn));memset(c, 0, sizeof(c));for (int i = 0; i < n; i++) {    for (int j = 0; j < n; j++) {scanf_(num);num %= 3;if (num == 0) continue;av[j][an[j]] = i;a[j][an[j]++] = num;    }}for (int i = 0; i < n; i++) {    for (int j = 0; j < n; j++) {scanf_(num);num %= 3;if (num == 0) continue;bv[i][bn[i]] = j;b[i][bn[i]++] = num;    }}for (int i = 0; i < n; i++) {    for (int j = 0; j < an[i]; j++) {for (int k = 0; k < bn[i]; k++) {    int x = av[i][j], y = bv[i][k];    c[x][y] = (c[x][y] + a[i][j] * b[i][k]) % 3;}    }}for (int i = 0; i < n; i++) {    for (int j = 0; j < n - 1; j++)printf("%d ", c[i][j]);    printf("%d\n", c[i][n - 1]);}    }    return 0;}

bitset：

#include <cstdio>#include <cstring>#include <string>#include <bitset>using namespace std;const int N = 805;int n, num;bitset<800> row[N][2], col[N][2];int main() {    while (~scanf("%d", &n)) {for (int i = 0; i < n; i++) {    row[i][0].reset();    row[i][1].reset();    col[i][0].reset();    col[i][1].reset();    for (int j = 0; j < n; j++) {scanf("%d", &num);if (num % 3 == 1)    row[i][0].set(j, 1);if (num % 3 == 2)    row[i][1].set(j, 1);    }}for (int i = 0; i < n; i++) {    for (int j = 0; j < n; j++) {scanf("%d", &num);if (num % 3 == 1)    col[j][0].set(i, 1);if (num % 3 == 2)    col[j][1].set(i, 1);    }}for (int i = 0; i < n; i++) {    for (int j = 0; j < n; j++) {int ans = 0;ans += (row[i][0]&col[j][0]).count();ans += 2 * (row[i][1]&col[j][0]).count() + 2 * (row[i][0]&col[j][1]).count();ans += (row[i][1]&col[j][1]).count();printf("%d%c", ans % 3, j == n - 1 ? '\n' : ' ');    }}    }    return 0;}

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