POJ3233 Matrix Power Series
Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.
Input
The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 109) and m (m < 104). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.
Output
Output the elements of S modulo m in the same way as A is given.
Sample Input
2 2 4 0 1 1 1
Sample Output
1 2 2 3
矩阵乘法 快速幂 递归
可以把矩阵当成矩阵里的元素,构建四个矩阵拼成的大矩阵进行快速幂递推;
也可以普通地递归计算:
项数为偶数的时候类似于 (A^1+A^2+A^3+A^4)*(A^4 + I) = A^1+A^2+A^3+A^4+A^5+A^6+A^7+A^7
项数为奇数的时候要加上那个奇数项
1 #include<iostream> 2 #include<algorithm> 3 #include<cstring> 4 #include<cstdio> 5 #include<cmath> 6 using namespace std; 7 int read(){ 8 int x=0,f=1;char ch=getchar(); 9 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 10 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} 11 return x*f; 12 } 13 int n,m,K; 14 struct Mat{ 15 int x[32][32]; 16 void init(){ 17 memset(x,0,sizeof x); 18 for(int i=1;i<=n;i++)x[i][i]=1; 19 return; 20 } 21 Mat operator + (const Mat &b){ 22 Mat res; 23 for(int i=1;i<=n;i++) 24 for(int j=1;j<=n;j++) 25 res.x[i][j]=(x[i][j]+b.x[i][j])%m; 26 return res; 27 } 28 Mat operator * (const Mat &b){ 29 Mat res; 30 for(int i=1;i<=n;i++) 31 for(int j=1;j<=n;j++){ 32 res.x[i][j]=0; 33 for(int k=1;k<=n;k++) 34 (res.x[i][j]+=x[i][k]*b.x[k][j]%m)%=m; 35 } 36 return res; 37 } 38 }I,A; 39 inline Mat ksm(Mat a,int k){ 40 Mat res;res.init(); 41 while(k){ 42 if(k&1)res=res*a; 43 a=a*a; 44 k>>=1; 45 } 46 return res; 47 } 48 Mat Bcalc(int k){ 49 if(k==1)return A; 50 if(k&1) 51 return (ksm(A,k>>1)+I)*Bcalc(k>>1)+ksm(A,k); 52 return (ksm(A,k>>1)+I)*Bcalc(k>>1); 53 } 54 int main(){ 55 int i,j; 56 n=read();K=read();m=read(); 57 I.init(); 58 for(i=1;i<=n;i++) 59 for(j=1;j<=n;j++) 60 A.x[i][j]=read(); 61 Mat ans=Bcalc(K); 62 for(i=1;i<=n;i++){ 63 for(j=1;j<=n;j++) 64 printf("%d ",ans.x[i][j]); 65 printf("\n"); 66 } 67 return 0; 68 }
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