HDU

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

Sample Output

2
2686

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
", ans%MOD);
    }
    return 0;
}