POJ 3233 Matrix Power Series 解题报告(子矩阵构造+矩阵快速幂)
2014-04-24 20:11
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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 Source POJ Monthly--2007.06.03, Huang, Jinsong |
解题报告: 求矩阵的n次方和。
如果接触过矩阵,都知道矩阵快速幂。一开始我也这么做的,二分计算前半段和后半段,即(A^1+A^2+...+A^(n/2))*(A^(n/2)+I)。题目是A了,但是耗时1600MS+。
看了一下Status,发现一堆0MS的……又想到整数的n次方和是有公式的,那么矩阵有吗?
百度了一下解题报告,发现了一个非常巧妙的方法:(引用:http://www.cnblogs.com/rainydays/archive/2011/02/21/1960189.html)
把问题转化以加速,令
B = A I
0 I
则B^(k + 1) = A^(k + 1) I + A + A2 + A3 + … + Ak
0 I
用二分法求B^(k + 1)
根据该性质,复杂度大大减小。另外矩阵相乘时每次最后再取模,会快上很多。
代码如下:
#include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; typedef long long LL; const int SIZE = 60; int n, mod; struct Matrix { int a[SIZE][SIZE]; Matrix(int t=0) { memset(a, 0, sizeof(a)); for(int i=0;i<SIZE;i++) a[i][i]=t; } Matrix operator*(const Matrix& b) const { Matrix c; for(int i=0;i<n;i++) for(int j=0;j<n;j++) { LL sum=0; for(int k=0;k<n;k++) sum+=a[i][k]*b.a[k][j]; c.a[i][j]=sum%mod; } return c; } }; Matrix pow(Matrix a, int b) { Matrix res(1); while(b) { if(b&1) res=res*a; a=a*a; b>>=1; } return res; } int main() { int k; while(~scanf("%d%d%d", &n, &k, &mod)) { Matrix a; for(int i=0;i<n;i++) for(int j=0;j<n;j++) scanf("%d", &a.a[i][j]), a.a[i][j]%=mod; for(int i=0;i<n;i++) a.a[i][i+n]=a.a[i+n][i+n]=1; n<<=1; a = pow(a, k+1); n>>=1; for(int i=0;i<n;i++) if(a.a[i][i+n]==0) a.a[i][i+n]=mod-1; else a.a[i][i+n]--; for(int i=0;i<n;i++, puts("")) for(int j=0;j<n;j++) printf("%d ", a.a[i][j+n]); } }
同样附上二分的代码:
#include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; int n, mod; struct Matrix { int a[30][30]; Matrix(int t=0) { memset(a, 0, sizeof(a)); for(int i=0;i<30;i++) a[i][i]=t; } Matrix operator*(const Matrix& b) const { Matrix c; for(int i=0;i<n;i++) for(int j=0;j<n;j++) for(int k=0;k<n;k++) c.a[i][j]=(c.a[i][j]+a[i][k]*b.a[k][j])%mod; return c; } Matrix operator+(const Matrix& b) const { Matrix c; for(int i=0;i<n;i++) for(int j=0;j<n;j++) c.a[i][j]=(a[i][j]+b.a[i][j])%mod; return c; } } one(1); Matrix pow(Matrix a, int b) { Matrix res(1); while(b) { if(b&1) res=res*a; a=a*a; b>>=1; } return res; } Matrix powSum(Matrix a, int b) { if(b==0) return one; else if(b==1) return a; else if(b%2==0) return powSum(a, b/2)*(pow(a, b/2)+one); else return powSum(a, b/2)*(pow(a, b/2)+one)+pow(a, b); } int main() { int k; while(~scanf("%d%d%d", &n, &k, &mod)) { Matrix a; for(int i=0;i<n;i++) for(int j=0;j<n;j++) scanf("%d", &a.a[i][j]), a.a[i][j]%=mod; a = powSum(a, k); for(int i=0;i<n;i++) for(int j=0;j<n;j++) printf("%d", a.a[i][j]%mod), printf(j==n-1?"\n":" "); } }
同样的,一个数的n次方和再模一个数也可以这样做。公式求法中需要除以一个数,而这个数在模mod系下不一定有逆元,所以用矩阵二分来做是个不错的方法。(和队友JX讨论的)。
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