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HDU 1575 Tr A 矩阵快速幂
跟着cxlove的矩阵专题(http://blog.csdn.net/ACM_cxlove/article/details/7815594)刷的,一道一道来。
最裸的题目,直接快速幂算就好了。
#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <climits>#include <iostream>#include <string>using namespace std; #define MP make_pair#define PB push_backtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int, int> PII;typedef pair<double, double> PDD;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 15;const int mod = 9973;struct Matrix { int n, m, data[maxn][maxn]; Matrix(int n = 0, int m = 0): n(n), m(m) { memset(data, 0, sizeof(data)); } void print() { for(int i = 1; i <= n; i++) { for(int j = 1; j <= m; j++) { printf("%d ", data[i][j]); } puts(""); } }};Matrix operator * (Matrix a, Matrix b) { int n = a.n, m = b.m; Matrix ret(n, m); for(int i = 1; i <= n; i++) { for(int j = 1; j <= m; j++) { for(int k = 1; k <= a.m; k++) { ret.data[i][j] += a.data[i][k] * b.data[k][j]; ret.data[i][j] %= mod; } } } return ret;}Matrix pow(Matrix mat, int k) { if(k == 1) return mat; Matrix ret = pow(mat * mat, k / 2); if(k & 1) ret = ret * mat; return ret;}Matrix mat;int n, k;int main() { int T; scanf("%d", &T); while(T--) { scanf("%d%d", &n, &k); mat.n = mat.m = n; for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { scanf("%d", &mat.data[i][j]); } } mat = pow(mat, k); int ans = 0; for(int i = 1; i <= n; i++) { ans = (ans + mat.data[i][i]) % mod; } printf("%d\n", ans); } return 0;}
HDU 1575 Tr A 矩阵快速幂
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