HDU 4920 Matrix multiplication(矩阵相乘)
2016-04-21 20:12
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各种TEL,233啊。没想到是处理掉0的情况就能够过啊。一直以为会有极端数据。没想到居然是这种啊、、在网上看到了一个AC的奇妙的代码,经典的矩阵乘法,仅仅只是把最内层的枚举,移到外面就过了啊、、、有点不理解啊,复杂度不是一样的吗、、
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 Aij. The next n lines describe the matrix B in similar format (0≤Aij,Bij≤109).
Output
For each tests:
Print n lines. Each of them contain n integers -- the matrix A×B in similar format.
Sample Input
Sample Output
Author
Xiaoxu Guo (ftiasch)
Source
2014 Multi-University Training Contest
5
这是看到有人交的AC的代码:
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 Aij. The next n lines describe the matrix B in similar format (0≤Aij,Bij≤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] %3); } } return 0; }
这是看到有人交的AC的代码:
#include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; const int N = 805; int a , b , ans ; 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++) //经典算法中这层循环在最内层。放最内层会超时,可是放在最外层或者中间都不会超时,不知道为什么 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; //假设在这里对3取余,就超时了 } for(i = 1; i <= n; i++) { for(j = 1; j < n; j++) printf("%d ", ans[i][j] % 3); printf("%d\n", ans[i] % 3); } } return 0; }
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