首页 > 代码库 > hdu 4965 矩阵快速幂 矩阵相乘性质

hdu 4965 矩阵快速幂 矩阵相乘性质

2024-07-18 11:02:54   230人阅读

Fast Matrix Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 170    Accepted Submission(s): 99

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 25 54 45 40 04 2 5 51 3 1 56 31 2 30 3 02 3 44 3 22 5 50 5 03 4 5 1 1 05 3 2 3 3 23 1 5 4 5 20 0
 
Sample Output
1456
 
Source
2014 Multi-University Training Contest 9
 
Recommend
hujie   |   We have carefully selected several similar problems for you:  4970 4968 4967 4966 4964 
 
题解：
 (4 <= N <= 1000)， (2 <=K <= 6)
N*K matrix A，K*N matrix B
A*B是N*N，但是B*A为k*k，于是。。。

 

  1 #include<iostream>  2 #include<cstring>  3 #include<cstdlib>  4 #include<cstdio>  5 #include<algorithm>  6 #include<cmath>  7 #include<queue>  8 #include<map>  9  10 #define N 1005 11 #define M 15 12 #define mod 6 13 #define mod2 100000000 14 #define ll long long 15 #define maxi(a,b) (a)>(b)? (a) : (b) 16 #define mini(a,b) (a)<(b)? (a) : (b) 17  18 using namespace std; 19  20 int n,k; 21 int a[N][10],b[10][N],d[10][10],f[N][10],g[N][N],h[N][N]; 22 int ans; 23  24 typedef  struct{ 25         int  m[10][10]; 26 }  Matrix; 27  28 Matrix e,P; 29  30 Matrix I = {1,0,0,0,0,0,0,0,0,0, 31             0,1,0,0,0,0,0,0,0,0, 32             0,0,1,0,0,0,0,0,0,0, 33             0,0,0,1,0,0,0,0,0,0, 34             0,0,0,0,1,0,0,0,0,0, 35             0,0,0,0,0,1,0,0,0,0, 36             0,0,0,0,0,0,1,0,0,0, 37             0,0,0,0,0,0,0,1,0,0, 38             0,0,0,0,0,0,0,0,1,0, 39             0,0,0,0,0,0,0,0,0,1, 40            }; 41  42 Matrix matrixmul(Matrix aa,Matrix bb) 43 { 44        int i,j,kk; 45        Matrix c; 46        for (i = 1 ; i <= k; i++) 47            for (j = 1; j <= k;j++) 48              { 49                  c.m[i][j] = 0; 50                  for (kk = 1; kk <= k; kk++) 51                      c.m[i][j] += (aa.m[i][kk] * bb.m[kk][j])%mod; 52                  c.m[i][j] %= mod; 53              } 54        return c; 55 } 56  57 Matrix quickpow(int num) 58 { 59        Matrix m = P, q = I; 60        while (num >= 1) 61        { 62              if (num & 1) 63                 q = matrixmul(q,m); 64              num = num >> 1; 65              m = matrixmul(m,m); 66        } 67        return q; 68 } 69  70 int main() 71 { 72     int i,j,o; 73     //freopen("data.in","r",stdin); 74     //scanf("%d",&T); 75     //for(int cnt=1;cnt<=T;cnt++) 76     //while(T--) 77     while(scanf("%d%d",&n,&k)!=EOF) 78     { 79         if(n==0 && k==0) break; 80         memset(d,0,sizeof(d)); 81         memset(f,0,sizeof(f)); 82         memset(g,0,sizeof(g)); 83         memset(h,0,sizeof(h)); 84         ans=0; 85         for(i=1;i<=n;i++){ 86             for(j=1;j<=k;j++){ 87                 scanf("%d",&a[i][j]); 88             } 89         } 90  91         for(i=1;i<=k;i++){ 92             for(j=1;j<=n;j++){ 93                 scanf("%d",&b[i][j]); 94             } 95         } 96  97         for(i=1;i<=k;i++){ 98             for(o=1;o<=k;o++){ 99                 for(j=1;j<=n;j++){100                     d[i][o]+=(b[i][j]*a[j][o])%6;101                 }102                 d[i][o]%=6;103                 P.m[i][o]=d[i][o];104             }105         }106 107 108 109         e=quickpow(n*n-1);110 111 112         for(i=1;i<=n;i++){113             for(o=1;o<=k;o++){114                 for(j=1;j<=k;j++){115                     f[i][o]+=(a[i][j]*e.m[j][o])%6;116                 }117                 f[i][o]%=6;118             }119         }120 121         for(i=1;i<=n;i++){122             for(o=1;o<=n;o++){123                 for(j=1;j<=k;j++){124                     g[i][o]+=(f[i][j]*b[j][o])%6;125                 }126                 g[i][o]%=6;127             }128         }129 /*130         for(i=1;i<=n;i++){131             for(o=1;o<=n;o++){132                 for(j=1;j<=n;j++){133                     h[i][o]+=(g[i][j]*g[j][o])%6;134                 }135                 h[i][o]%=6;136             }137         }138 139 */140 141         for(i=1;i<=n;i++){142             for(o=1;o<=n;o++){143                 ans+=g[i][o];144             }145         }146         printf("%d\n",ans);147 148     }149 150     return 0;151 }


 


联系
我们