HDU - problem Tr A 【快速幂 + 矩阵】

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Tr A

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4488 Accepted Submission(s): 3377

Problem Description

A为一个方阵，则Tr A表示A的迹（就是主对角线上各项的和），现要求Tr(A^k)%9973。

Input

数据的第一行是一个T，表示有T组数据。
每组数据的第一行有n(2 <= n <= 10)和k(2 <= k < 10^9)两个数据。接下来有n行，每行有n个数据，每个数据的范围是[0,9]，表示方阵A的内容。

Output

对应每组数据，输出Tr(A^k)%9973。

Sample Input

22 21 00 13 999999991 2 34 5 67 8 9

Sample Output

Author

xhd

Source

HDU 2007-1 Programming Contest

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#include <map>#include <set>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <iostream>#include <stack>#include <cmath>#include <string>#include <vector>#include <cstdlib>//#include <bits/stdc++.h>//#define LOACL#define space " "using namespace std;typedef long long LL;typedef __int64 Int;typedef pair<int, int> paii;const int INF = 0x3f3f3f3f;const double ESP = 1e-5;const double PI = acos(-1.0);const int MOD = 9973;const int MAXN = 15;int n, m;struct Matrix {LL m[MAXN][MAXN];int row, col;};Matrix ori, res, u;void init() {memset(res.m, 0, sizeof(res.m));ori.row = ori.col = n;}void scan_in() {for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {scanf("%d", &ori.m[i][j]);}}}Matrix multi(Matrix x, Matrix y) {Matrix z;memset(z.m, 0, sizeof(z.m));z.row = x.row; z.col = y.col;for (int i = 1; i <= x.row; i++) {for (int k = 1; k <= x.col; k++) {for (int j = 1; j <= y.col; j++) {z.m[i][j] += x.m[i][k]*y.m[k][j]%MOD;}z.m[i][k] %= MOD;}}return z;}Matrix pow_mod(Matrix a, int x){Matrix b;b.col = a.col; b.row = a.row;memset(b.m, 0, sizeof(b.m));for (int i = 1; i <= n; i++) {b.m[i][i] = 1;}    while(x){        if(x&1) b = multi(a,b);        a = multi(a, a);        x >>= 1;    }    return b;}int main() {int T;scanf("%d", &T);while (T--) {scanf("%d%d", &n, &m);init(); scan_in();res = pow_mod(ori, m);int ans = 0;for (int i = 1; i <= n; i++) {ans += res.m[i][i];}printf("%d\n", ans%MOD);}return 0;}

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