hdu4920 矩阵相乘

 Given two matrices A and B of size n×n, find the product of them. bobo hates big integers. So you are only asked to find the result modulo 3. Input The input consists of several tests. For each tests: The first line contains n (1≤n≤800). Each of the following n lines contain n integers -- the description of the matrix A. The j-th integer in the i-th line equals A _ij. The next n lines describe the matrix B in similar format (0≤A _ij,B _ij≤10 ⁹). Output For each tests: Print n lines. Each of them contain n integers -- the matrix A×B in similar format. Sample Input 1 0 1 2 0 1 2 3 4 5 6 7 Sample Output 0 0 1 2 1   给你两个矩阵，然后让你计算矩阵的乘积 直接乘就可以了，有0的话直接跳过不要乘，还有不要把取余运算放在三重循环中（会超时） ac代码： #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> using namespace std; int a[1000][1000],b[1000][1000],c[1000][1000]; int main() { int n; while(scanf("%d",&n)!=-1) { for(int i=0; i<n; i++) for(int j=0; j<n; j++) { scanf("%d",&a[i][j]); a[i][j]=a[i][j]%3; c[i][j]=0; } for(int i=0; i<n; i++) for(int j=0; j<n; j++) { scanf("%d",&b[i][j]); b[i][j]=b[i][j]%3; } for(int i=0; i<n; i++) for(int k=0; k<n; k++) if(a[i][k]) for(int j=0; j<n; j++) c[i][j]=(c[i][j]+a[i][k]*b[k][j])%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; }