POJ 3233 Matrix Power Series (矩阵&快速等比数列求和取模)
2013-08-25 20:30
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Matrix Power Series
http://poj.org/problem?id=3233
Time Limit: 3000MS
Memory Limit: 131072K
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
Sample Output
题意要求矩阵S=A+A^2+A^3+...+A^k mod m,可以用二分的方法。
单个数的快速幂请参考这篇文章。
快速等比数列求和:如何求k=19的情况呢?首先根据19的二进制10011构造这么一阶差分序列:1,2,16,所以原序列是0,1,3,19.
那怎么求A+A^2+A^3+...+A^19呢?我们就按照原序列来求,也就要先求一阶差分序列。
由于第一个数是0,我们初始化ans=0
第二个数是1,ans=ans*A^1+A=A
第三个数是3,ans=ans*A^2+A(I+A)=A+A^2+A^3
由于二进制中间有两个0,所以与ans相乘的数变为A^16(算法:A^2*A^2=A^4,A^4*A^4=A^16)
与ans*A^16相加的数变为A+...A^16(算法:(A+A^2)*((I+A^2)=A+...+A^4,(A+...+A^4)*(I+A^4)=A+...+A^8,(A+...+A^8)*(I+A^8)=A+...A^16)
第四个数是19,ans=ans*A^16+(A+...A^16)
完整代码:
http://poj.org/problem?id=3233
Time Limit: 3000MS
Memory Limit: 131072K
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
题意要求矩阵S=A+A^2+A^3+...+A^k mod m,可以用二分的方法。
单个数的快速幂请参考这篇文章。
快速等比数列求和:如何求k=19的情况呢?首先根据19的二进制10011构造这么一阶差分序列:1,2,16,所以原序列是0,1,3,19.
那怎么求A+A^2+A^3+...+A^19呢?我们就按照原序列来求,也就要先求一阶差分序列。
由于第一个数是0,我们初始化ans=0
第二个数是1,ans=ans*A^1+A=A
第三个数是3,ans=ans*A^2+A(I+A)=A+A^2+A^3
由于二进制中间有两个0,所以与ans相乘的数变为A^16(算法:A^2*A^2=A^4,A^4*A^4=A^16)
与ans*A^16相加的数变为A+...A^16(算法:(A+A^2)*((I+A^2)=A+...+A^4,(A+...+A^4)*(I+A^4)=A+...+A^8,(A+...+A^8)*(I+A^8)=A+...A^16)
第四个数是19,ans=ans*A^16+(A+...A^16)
完整代码:
/*79ms,144KB*/ #include <cstdio> #include <cstring> const int matSize = 31; int calSize = matSize, mod = 1000000007; struct Matrix { int val[matSize][matSize]; Matrix(bool Init = false)///使用默认实参以减少代码量 { for (int i = 0; i < calSize; i++) { for (int j = 0; j < calSize; j++) val[i][j] = 0; if (Init) val[i][i] = 1;///单位矩阵 } } void print() { for (int i = 0; i < calSize; i++) { for (int j = 0; j < calSize; j++) { if (j) putchar(' '); printf("%d", val[i][j]); } puts(""); } } } Base; Matrix operator + (Matrix &_a, Matrix &_b) { Matrix ret; for (int i = 0; i < calSize; i++) for (int j = 0; j < calSize; j++) ret.val[i][j] = (_a.val[i][j] + _b.val[i][j]) % mod; return ret; } Matrix operator * (Matrix &_a, Matrix &_b) { Matrix ret = Matrix(); for (int i = 0; i < calSize; i++) for (int k = 0; k < calSize; k++) if (_a.val[i][k])///优化一下 for (int j = 0; j < calSize; j++) { ret.val[i][j] += _a.val[i][k] * _b.val[k][j]; ret.val[i][j] %= mod; } return ret; } void deal(int k) { Matrix one = Matrix(true), ans = Matrix(); Matrix ep = Base, tmp = Base, temp; while (k) { if (k & 1) { ans = ans * ep; ans = ans + tmp; } temp = one + ep; tmp = tmp * temp; ep = ep * ep; k >>= 1; } ans.print(); } int main() { int k; while (~scanf("%d%d%d", &calSize, &k, &mod)) { for (int i = 0; i < calSize; i++) for (int j = 0; j < calSize; j++) { scanf("%d", &Base.val[i][j]); Base.val[i][j] %= mod; } deal(k); } return 0; }
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