各種TEL，233啊。沒想到是處理掉0的情況就能夠過啊。一直以為會有極端數據。沒想到居然是這種啊、、在網上看到了一個AC的奇妙的代碼，經典的矩陣乘法，僅僅只是把最內層的枚舉，移到外面就過了啊、、、有點不理解啊，復雜度不是一樣的嗎、、

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)????Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 640????Accepted Submission(s): 250


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 1 0 1 2 0 1 2 3 4 5 6 7
Sample Output 0 0 1 2 1
Author Xiaoxu Guo (ftiasch)
Source 2014 Multi-University Training Contest 5 #include <algorithm> #include <iostream> #include <stdlib.h> #include <string.h> #include <iomanip> #include <stdio.h> #include <string> #include <queue> #include <cmath> #include <stack> #include <map> #include <set> #define eps 1e-12 ///#define M 1000100 #define LL __int64 ///#define LL long long ///#define INF 0x7ffffff #define INF 0x3f3f3f3f #define PI 3.1415926535898 #define zero(x) ((fabs(x)<eps)?

0:x) using namespace std; const int maxn = 810; int a[maxn][maxn]; int b[maxn][maxn]; int c[maxn][maxn]; int aa[maxn][maxn]; int bb[maxn][maxn]; int main() { int n; while(cin >>n) { memset(c, 0, sizeof(c)); memset(aa, 0, sizeof(aa)); memset(bb, 0, sizeof(bb)); for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { scanf("%d",&a[i][j]); a[i][j] %= 3; } } for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { scanf("%d",&b[i][j]); b[i][j] %= 3; } } for(int i = 1; i <= n; i++) { int x = -1; for(int j = n; j >= 0; j--) { aa[i][j] = x; if(a[i][j]) x = j; } } for(int i = 1; i <= n; i++) { int x = -1; for(int j = n; j >= 0; j--) { bb[i][j] = x; if(b[i][j]) x = j; } } for (int i = 1; i <= n; i++) { for(int j = aa[i][0]; j != -1; j = aa[i][j]) { for(int k = bb[j][0]; k != -1; k = bb[j][k]) c[i][k] += a[i][j]*b[j][k]; } } for(int i = 1; i <= n; i++) { for(int j = 1; j <= n-1; j++) printf("%d ",c[i][j]%3); printf("%d\n",c[i][n]%3); } } return 0; }

這是看到有人交的AC的代碼：

#include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; const int N = 805; int a[N][N], b[N][N], ans[N][N]; int main() {int n, i, j, k;while(~scanf("%d",&n)){for(i = 1; i <= n; i++)for(j = 1; j <= n; j++){scanf("%d",&a[i][j]);a[i][j] %= 3;}for(i = 1; i <= n; i++)for(j = 1; j <= n; j++){scanf("%d",&b[i][j]);b[i][j] %= 3;}memset(ans, 0, sizeof(ans));for(k = 1; k <= n; k++) //經典算法中這層循環在最內層。放最內層會超時，可是放在最外層或者中間都不會超時，不知道為什么for(i = 1; i <= n; i++)for(j = 1; j <= n; j++){ans[i][j] += a[i][k] * b[k][j];//ans[i][j] %= 3; //假設在這里對3取余，就超時了}for(i = 1; i <= n; i++){for(j = 1; j < n; j++)printf("%d ", ans[i][j] % 3);printf("%d\n", ans[i][n] % 3);}}return 0; }

轉載于:https://www.cnblogs.com/yxwkf/p/5418535.html

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