Matrix Power Series
Time Limit: 3000MS | Memory Limit: 131072K | |
Total Submissions: 11954 | Accepted: 5105 |
Description
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
Source
POJ Monthly--2007.06.03, Huang, Jinsong
题意:给出一个n*n的矩阵,和一个正整数k,S=A+A^2+A^3+......+A^k;矩阵中的每一个元素对m取模
当k=1的时候,f(1)=A;当k=2时 f(2)=A+A^2,当k=3时f(3)=A+A^2+A^3 = A+A*(A+A^2)=A+A*( f(2) );所以我们得出 f(k)=f(k-1)*A+A;
然后我们就可以试着构建矩阵:| A 1 | *| f(k-1) | =| f(k) |,再进行矩阵快速幂就好啦。
| 0 1 | | A | | A |
#include<algorithm> #include<iostream> #include<stdio.h> using namespace std; int n,m,k; struct Matrix { long long val[30][30]; }; Matrix operator*(const Matrix &a,const Matrix &b) { int i,j,k; Matrix ans; for(i=0;i<n;i++) for(j=0;j<n;j++) ans.val[i][j]=0; for(i=0;i<n;i++) { for(j=0;j<n;j++) { for(k=0;k<n;k++) { ans.val[i][j]+=a.val[i][k]*b.val[k][j]; } ans.val[i][j]%=m; } } return ans; } Matrix operator+(const Matrix &a,const Matrix &b) { int i,j,k; Matrix ans; for(i=0;i<n;i++) for(j=0;j<n;j++) ans.val[i][j]=a.val[i][j]+b.val[i][j],ans.val[i][j]%=m; return ans; } struct Matrix_1 { Matrix num[2][2]; }; Matrix_1 operator*(const Matrix_1 &a,const Matrix_1 &b) { int i,j,k; Matrix_1 ans; Matrix p; for(i=0;i<n;i++) { for(j=0;j<n;j++) { p.val[i][j]=0; } } ans.num[0][0]=ans.num[0][1]=ans.num[1][0]=ans.num[1][1]=p; for(i=0;i<2;i++) { for(j=0;j<2;j++) { for(k=0;k<2;k++) { ans.num[i][j]=ans.num[i][j]+a.num[i][k]*b.num[k][j]; } } } return ans; } void solve(int b,Matrix f) { Matrix p,q; Matrix_1 ans,a; int i,j,k; for(i=0;i<n;i++) { for(j=0;j<n;j++) { if(i==j) p.val[i][j]=1;//单位矩阵 else p.val[i][j]=0; q.val[i][j]=0;//0矩阵 } } ans.num[0][0]=ans.num[1][1]=p; ans.num[0][1]=ans.num[1][0]=q; a.num[0][0]=f; a.num[0][1]=a.num[1][1]=p; a.num[1][0]=q; while(b) { if(b&1) ans=ans*a; b>>=1; a=a*a; } f=ans.num[0][0]*f+ans.num[0][1]*f; for(int i=0;i<n;i++) { for(int j=0;j<n;j++) { printf("%lld ",f.val[i][j]%m); } printf("\n"); } } int main() { int i,j,k; Matrix p; cin>>n>>k>>m; for(i=0;i<n;i++) { for(j=0;j<n;j++) { scanf("%lld",&p.val[i][j]); } } solve(k-1,p); }