HDU 4920 稀疏矩阵相乘

标签：io   for   amp   size   algorithm   hdu   return   on   

暴力啊，不要想太多

T^T g艹太坑爹了

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

using namespace std;

const int MAX_N = 807;

int n;
int a[MAX_N][MAX_N], b[MAX_N][MAX_N];
int c[MAX_N][MAX_N];

int main() {
    while (1 == scanf("%d", &n)) {
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                int x;
                scanf("%d", &x);
                a[i][j] = x % 3;
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                int x;
                scanf("%d", &x);
                b[i][j] = x % 3;
            }
        }
        memset(c, 0, sizeof(c));
        for (int i = 0; i < n; ++i) {
            for (int k = 0; k < n; ++k) {
                if (a[i][k] == 0) continue;
                for (int j = 0; j < n; ++j) {
                    c[i][j] += a[i][k] * b[k][j];
                    c[i][j] %= 3;
                }
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (j == 0) printf("%d", c[i][j]);
                else printf(" %d", c[i][j]);
            }
            puts("");
        }
    }
    return 0;
}

HDU 4920 稀疏矩阵相乘,布布扣,bubuko.com

HDU 4920 稀疏矩阵相乘

标签：io   for   amp   size   algorithm   hdu   return   on   

原文地址：http://blog.csdn.net/qq574857122/article/details/38388419