HDU-4965 Fast Matrix Calculation （矩阵快速幂）
Fast Matrix Calculation
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.
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.
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
看到矩阵快速幂以为是一道模板水题（其实就是一道模板水题。。。），就欢快的找模板了，发现n能取1000时，果断CE（栈内存爆了）+TLE，然后就不知所措
最后队友将C^(n*n)写成(A*B)*(A*B)……才反应过来，利用矩阵点结合律，计算(B*A)^(n*n-1)就行了，思维太死了。。。
#include <cstdio>
#include <cstring>
using namespace std;
int n,k,ans;
const int MAXN=1001;
const int MOD=6;
struct Matrix {
int x[MAXN][MAXN];
void clear() {
memset(x,0,sizeof(x));
};
}a,b,c,tmp,pro;
void mul(const Matrix& x,const Matrix& y,int bound_1,int bound_2,int bound_3) {//刚开始写了个带优化O(n^3)的乘法，为了减少更改其他代码，故参数多加了3个循环变量点上界以使其适用于所有情况
tmp.clear();
for(int i=0;i<bound_1;++i)
for(int j=0;j<bound_2;++j)
for(int k=0;k<bound_3;++k)
tmp.x[i][j]=(tmp.x[i][j]+x.x[i][k]*y.x[k][j])%MOD;
}
void quick_pow(Matrix& x,int num) {//矩阵快速幂
pro.clear();
for(int i=0;i<k;++i)
pro.x[i][i]=1;
while(num>0) {
if((num&1)==1) {
mul(pro,x,k,k,k);
pro=tmp;
}
mul(x,x,k,k,k);
x=tmp;
num>>=1;
}
}
int main() {
while(scanf("%d%d",&n,&k),n!=0&&k!=0) {
a.clear();
b.clear();
c.clear();
for(int i=0;i<n;++i) {
for(int j=0;j<k;++j)
scanf("%d",&a.x[i][j]);
}
for(int i=0;i<k;++i) {
for(int j=0;j<n;++j) {
scanf("%d",&b.x[i][j]);
}
}
mul(b,a,k,k,n);//由于矩阵太大，函数返回矩阵的话oj直接判CE，只能用全局变量
c=tmp;
quick_pow(c,n*n-1);
c=pro;
mul(a,c,n,n,k);
c=tmp;
mul(c,b,n,n,k);
c=tmp;
ans=0;
for(int i=0;i<n;++i)
for(int j=0;j<n;++j)
ans+=c.x[i][j];
printf("%d\n",ans);
}
return 0;
}
看了题解后发现，只在快速幂时用结构体，其余时候用二位数组，手动乘法，可以减少大量点空间浪费和赋值点时间浪费
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=6;
const int MOD=6;
int n,m,ans;
int a[1001][1001],b[1001][1001],c[1001][1001];
struct Matrix {
int x[MAXN][MAXN];
Matrix() {
memset(x,0,sizeof(x));
}
void clear() {
memset(x,0,sizeof(x));
};
}cur;
Matrix mul(const Matrix& x,const Matrix& y) {
Matrix res;
for(int i=0;i<MAXN;++i)
for(int j=0;j<MAXN;++j)
for(int k=0;k<MAXN;++k)
res.x[i][j]=(res.x[i][j]+x.x[i][k]*y.x[k][j])%MOD;
return res;
}
Matrix quick_pow(Matrix& x,int num) {//矩阵快速幂
Matrix res;
for(int i=0;i<MAXN;++i)
res.x[i][i]=1;
while(num>0) {
if((num&1)==1)
res=mul(res,x);
x=mul(x,x);
num>>=1;
}
return res;
}
int main() {
while(scanf("%d%d",&n,&m),n!=0&&m!=0) {
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]);
cur.clear();
for(int i=0;i<m;++i)
for(int j=0;j<m;++j)
for(int k=0;k<n;++k)
cur.x[i][j]=(cur.x[i][j]+b[i][k]*a[k][j])%MOD;
cur=quick_pow(cur,n*n-1);
memset(c,0,sizeof(c));
for(int i=0;i<n;++i)
for(int j=0;j<n;++j)
for(int k=0;k<m;++k)
c[i][j]=(c[i][j]+a[i][k]*cur.x[k][j])%MOD;
memset(a,0,sizeof(a));
for(int i=0;i<n;++i)
for(int j=0;j<n;++j)
for(int k=0;k<m;++k)
a[i][j]=(a[i][j]+c[i][k]*b[k][j])%MOD;
ans=0;
for(int i=0;i<n;++i)
for(int j=0;j<n;++j)
ans+=a[i][j];
printf("%d\n",ans);
}
return 0;
}