hdu 4920 Matrix multiplication(高效)

题目链接：4920 Matrix multiplication

题目大意：给定两个n阶矩阵，求矩阵相乘后模3.

解题思路：因为矩阵模掉3后只有0，1，2三种情况。所以对于矩阵A，记录每一行中1，2的位置，借助bitset。矩阵B中每一列1，2的位置。然后对于结果中每个位置，只要考虑1?1,1?2,2?1，2?2的个数即可。

#include <cstdio>
#include <cstring>
#include <bitset>
#include <algorithm>

using namespace std;
const int maxn = 805;

int N, C[maxn][maxn];
bitset<maxn> 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<maxn> 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;
}

hdu 4920 Matrix multiplication(高效),布布扣,bubuko.com

hdu 4920 Matrix multiplication(高效)

原文：http://blog.csdn.net/keshuai19940722/article/details/38391913