POJ -- 3233 求“等比矩阵”前n(n <=10^9)项和
Matrix Power Series
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
思路:1.最基本的,需要用到矩阵快速幂 2.快速幂求完之后怎样快速求和?若逐项累加求和必然会超时,这时需要求递推公式:(1)若n为偶数,则:S(n) = A^(n/2)*S(n/2)+s(n/2);(2)若n为奇数 S(n) = A^(n/2+1) + S(n/2)*A^(n/2+1) + S(n/2),公式不难推,写几个就发现规律了。这样就把时间复杂度降下来了。
1 #include<cstdio> 2 #include<string> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 int n, m; 7 typedef struct Matrix{ 8 int m[30][30]; 9 Matrix(){ 10 memset(m, 0, sizeof(m)); 11 } 12 }Matrix; 13 Matrix mtAdd(Matrix A, Matrix B){ 14 for(int i = 0;i < n;i ++) 15 for(int j = 0;j < n;j ++){ 16 A.m[i][j] += B.m[i][j]; 17 A.m[i][j] %= m; 18 } 19 return A; 20 } 21 Matrix mtMul(Matrix A, Matrix B){ 22 Matrix tmp; 23 for(int i = 0;i < n;i ++) 24 for(int j = 0;j < n;j ++) 25 for(int k = 0;k < n;k ++){ 26 tmp.m[i][j] += A.m[i][k]*B.m[k][j]; 27 tmp.m[i][j] %= m; 28 } 29 return tmp; 30 } 31 Matrix mtPow(Matrix A, int k){ 32 if(k == 1) return A; 33 Matrix tmp = mtPow(A, k >> 1); 34 Matrix res = mtMul(tmp, tmp); 35 if(k&1) res = mtMul(res, A); 36 return res; 37 } 38 Matrix mtSum(Matrix A, int k){ 39 if(k == 1) return A; 40 Matrix tmp = mtSum(A, k/2); 41 if(k&1){ 42 Matrix t = mtPow(A, k/2+1); 43 Matrix tmp1 = mtMul(tmp, t); 44 Matrix tmp2 = mtAdd(t, tmp); 45 return mtAdd(tmp1, tmp2); 46 }else return mtAdd(tmp, mtMul(mtPow(A, k/2), tmp)); 47 } 48 int main(){ 49 int k, tmp; 50 /* freopen("in.c", "r", stdin); */ 51 while(~scanf("%d%d%d", &n, &k, &m)){ 52 Matrix M; 53 for(int i = 0;i < n;i ++) 54 for(int j = 0;j < n;j ++){ 55 scanf("%d", &tmp); 56 M.m[i][j] = tmp; 57 } 58 M = mtSum(M, k); 59 for(int i = 0;i < n;i ++){ 60 for(int j = 0;j < n;j ++) 61 printf("%d ", M.m[i][j]); 62 puts(""); 63 } 64 } 65 return 0; 66 }