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.
矩阵快速幂
1 #include<stdio.h>
2 #include<string.h>
3 #include<math.h>
4 typedef long long ll;
5 const int mod=6;
6
7 struct mat{
8 int r,c;
9 int m[7][7]; //经测试最大开成590*590的 ll 型矩阵
10 mat(){}
11 mat(int r,int c):r(r),c(c){}
12 void clear(){
13 memset(m,0,sizeof(m));
14 }
15
16 mat operator+(mat a)const{
17 mat ans(r,c);
18 for(int i=1;i<=r;i++){
19 for(int j=1;j<=c;j++){
20 ans.m[i][j]=(m[i][j]+a.m[i][j])%mod;
21 }
22 }
23 return ans;
24 }
25
26 mat operator*(mat a)const{
27 mat tmp(r,a.c);
28 int i,j,k;
29 for(i=1;i<=tmp.r;i++){
30 for(j=1;j<=tmp.c;j++){
31 tmp.m[i][j]=0;
32 for(k=1;k<=c;k++){
33 tmp.m[i][j]=(tmp.m[i][j]+(m[i][k]*a.m[k][j])%mod)%mod;
34 }
35 }
36 }
37 return tmp;
38 }
39
40 mat operator^(int n)const{ //需要时可以用 ll n，注意运算符优先级比较低，多用括号；
41 mat ans(r,r),tmp(r,r);
42 memcpy(tmp.m,m,sizeof(tmp.m));
43 ans.clear();
44 for(int i=1;i<=ans.r;i++){
45 ans.m[i][i]=1;
46 }
47 while(n){
48 if(n&1)ans=ans*tmp;
49 n>>=1;
50 tmp=tmp*tmp;
51 }
52 return ans;
53 }
54
55 void print()const{
56 for(int i=1;i<=r;i++){
57 for(int j=1;j<=c;j++){
58 printf("%d",m[i][j]);
59 if(j==c)printf("
");
60 else printf(" ");
61 }
62 }
63 }
64
65 };
66
67 int m1[2000][2000],m2[2000][2000],tmp[7][7],tmp2[2000][2000],tmp3[2000][2000];
68
69 int main(){
70 int n,k;
71 while(scanf("%d%d",&n,&k)!=EOF&&n+k){
72 int i,j,p;
73 for(i=1;i<=n;i++){
74 for(j=1;j<=k;j++)scanf("%d",&m1[i][j]);
75 }
76 for(i=1;i<=k;i++){
77 for(j=1;j<=n;j++)scanf("%d",&m2[i][j]);
78 }
79 for(i=1;i<=k;i++){
80 for(j=1;j<=k;j++){
81 tmp[i][j]=0;
82 for(p=1;p<=n;p++){
83 tmp[i][j]+=m2[i][p]*m1[p][j];
84 }
85 tmp[i][j]%=6;
86 }
87 }
88 mat a(k,k);
89 memcpy(a.m,tmp,sizeof(tmp));
90 a=(a^(n*n-1));
91 memcpy(tmp,a.m,sizeof(tmp));
92 for(i=1;i<=n;i++){
93 for(j=1;j<=k;j++){
94 tmp2[i][j]=0;
95 for(p=1;p<=k;p++){
96 tmp2[i][j]+=m1[i][p]*tmp[p][j];
97 }
98 tmp2[i][j]%=6;
99 }
100 }
101 for(i=1;i<=n;i++){
102 for(j=1;j<=n;j++){
103 tmp3[i][j]=0;
104 for(p=1;p<=k;p++){
105 tmp3[i][j]+=tmp2[i][p]*m2[p][j];
106 }
107 tmp3[i][j]%=6;
108 }
109 }
110 int ans=0;
111 for(i=1;i<=n;i++){
112 for(j=1;j<=n;j++)ans+=tmp3[i][j];
113 }
114 printf("%d
",ans);
115 }
116 return 0;
117 }