HDU 4920(Matrix multiplication-矩阵乘法优化)

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3647    Accepted Submission(s): 1522

Problem Description

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≤10⁹).

 

Output

For each tests:

Print n lines. Each of them contain n integers -- the matrix A×B in similar format.

 

Sample Input

10120 12 34 56 7

 

Sample Output

00 12 1

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

 

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直接矩阵乘 O(n^3) 800^3=5120,0000 TLE

所以利用mod3的性质 稍加优化

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)|#define MAXN (800+10)#define F (3)#define pb push_back#define mp make_pair typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int n,a[MAXN][MAXN],b[MAXN][MAXN],c[MAXN][MAXN];int main(){//freopen("j.in","r",stdin);while(cin>>n) {For(i,n) For(j,n) scanf("%d",&a[i][j]),a[i][j]%=3;For(i,n) For(j,n) scanf("%d",&b[i][j]),b[i][j]%=3,c[i][j]=0;For(i,n) For(k,n)if (a[i][k]) For(j,n) c[i][j]=(c[i][j]+a[i][k]*b[k][j])%3;For(i,n) {For(j,n-1) printf("%d ",c[i][j]);printf("%d\n",c[i][n]);}}return 0;}