2014多校聯合第9場1006||hdu 4965 矩陣乘法和快速冪

life4711發表於2014-08-21
原文網址 : https://blog.csdn.net/lvshubao1314/article/details/38725389
矩陣

http://acm.hdu.edu.cn/showproblem.php?pid=4965

Problem Description
One day, Alice and Bob felt bored again, Bob knows Alice is a girl who loves math and is just learning something about matrix, so he decided to make a crazy problem for her.

Bob has a six-faced dice which has numbers 0, 1, 2, 3, 4 and 5 on each face. At first, he will choose a number N (4 <= N <= 1000), and for N times, he keeps throwing his dice for K times (2 <=K <= 6) and writes down its number on the top face to make an N*K matrix A, in which each element is not less than 0 and not greater than 5. Then he does similar thing again with a bit difference: he keeps throwing his dice for N times and each time repeat it for K times to write down a K*N matrix B, in which each element is not less than 0 and not greater than 5. With the two matrix A and B formed, Alice’s task is to perform the following 4-step calculation.

Step 1: Calculate a new N*N matrix C = A*B.
Step 2: Calculate M = C^(N*N). 
Step 3: For each element x in M, calculate x % 6. All the remainders form a new matrix M’.
Step 4: Calculate the sum of all the elements in M’. 

Bob just made this problem for kidding but he sees Alice taking it serious, so he also wonders what the answer is. And then Bob turn to you for help because he is not good at math.
 

Input
The input contains several test cases. Each test case starts with two integer N and K, indicating the numbers N and K described above. Then N lines follow, and each line has K integers between 0 and 5, representing matrix A. Then K lines follow, and each line has N integers between 0 and 5, representing matrix B.

The end of input is indicated by N = K = 0.
 

Output
For each case, output the sum of all the elements in M’ in a line.
 

Sample Input
4 2
5 5
4 4
5 4
0 0
4 2 5 5
1 3 1 5
6 3
1 2 3
0 3 0
2 3 4
4 3 2
2 5 5
0 5 0
3 4 5 1 1 0
5 3 2 3 3 2
3 1 5 4 5 2
0 0





 



Sample Output




14
56





矩陣題。請看官方題解






#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
int a[1005][6],b[6][1005],c[6][6],s[1005][1005],sum[1005][6];
const int MAX=6;
int n,m;
struct Matrix
{
    int m[MAX][MAX];
};
Matrix I={1,0,0,0,0,0,
          0,1,0,0,0,0,
          0,0,1,0,0,0,
          0,0,0,1,0,0,
          0,0,0,0,1,0,
          0,0,0,0,0,1};
Matrix matrixmul(Matrix a,Matrix b)
{
    int i,j,k;
    Matrix c;
    for(i=0;i<m;i++)
        for(j=0;j<m;j++)
        {
           c.m[i][j]=0;
           for(k=0;k<m;k++)
            c.m[i][j]+=(a.m[i][k]*b.m[k][j])%6;
           c.m[i][j]%=6;
        }
    return c;
}
Matrix quickpow(Matrix P,int n)
{
    Matrix m=P,b=I;
    while(n>=1)
    {
        if(n&1)
            b=matrixmul(b,m);
        n=n>>1;
        m=matrixmul(m,m);
    }
    return b;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)
            break;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(s,0,sizeof(s));
        memset(c,0,sizeof(c));
        memset(sum,0,sizeof(sum));
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
                 scanf("%d",&a[i][j]);
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
                 scanf("%d",&b[i][j]);
        for(int i=0;i<m;i++)
            for(int l=0;l<m;l++)
            {
               for(int j=0;j<n;j++)
                  c[i][l]+=b[i][j]*a[j][l];
            }
       /* for(int i=0;i<m;i++)
        {
            for(int j=0;j<m;j++)
                printf("%d ",c[i][j]);
            printf("\n");
        }*/
        Matrix A;
        for(int i=0;i<m;i++)
            for(int j=0;j<m;j++)
                A.m[i][j]=c[i][j];
        /* for(int i=0;i<m;i++)
        {
            for(int j=0;j<m;j++)
                printf("%d ",A.m[i][j]);
            printf("\n");
        }*/
        A=quickpow(A,n*n-1);

       /* printf("\n");
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<m;j++)
                printf("%d ",A.m[i][j]);
            printf("\n");
        }
        printf("\n");
*/
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
                for(int k=0;k<m;k++)
                    sum[i][j]+=a[i][k]*A.m[k][j];
        for(int i=0;i<n;i++)
            for(int k=0;k<n;k++)
            {
                for(int j=0;j<m;j++)
                     s[i][k]+=sum[i][j]*b[j][k];
                 s[i][k]%=6;
            }
        int ans=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                ans+=s[i][j];
        printf("%d\n",ans);
    }
    return 0;
}



            




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