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HDU 1575 Tr A(矩阵快速幂)

2016-03-05 00:23 513 查看
题意:

矩阵快速幂模板。

代码:

[code]//
//  Created by  CQU_CST_WuErli
//  Copyright (c) 2016 CQU_CST_WuErli. All rights reserved.
//
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <queue>
#include <stack>
#include <set>
#include <algorithm>
#include <sstream>
#define CLR(x) memset(x,0,sizeof(x))
#define OFF(x) memset(x,-1,sizeof(x))
#define MEM(x,a) memset((x),(a),sizeof(x))
#define BUG cout << "I am here" << endl
#define lookln(x) cout << #x << "=" << x << endl
#define SI(a) scanf("%d",&a)
#define SII(a,b) scanf("%d%d",&a,&b)
#define SIII(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define rep(flag,start,end) for(int flag=start;flag<=end;flag++)
#define Rep(flag,start,end) for(int flag=start;flag>=end;flag--)
#define Lson l,mid,rt<<1
#define Rson mid+1,r,rt<<1|1
#define Root 1,n,1
const int INF_INT=0x3f3f3f3f;
const long long INF_LL=0x7fffffff;
const int MOD=1e9+7;
const double eps=1e-9;
const double pi=acos(-1);
typedef long long  ll;
using namespace std;

int mod=9973;
int n,k;
struct Matrix {
    int mat[11][11];
}initM,ansM;

Matrix multi(Matrix& a,Matrix& b) {
    Matrix ret;
    rep(i,0,n-1) rep(j,0,n-1) {
        ret.mat[i][j]=0;
        rep(k,0,n-1) ret.mat[i][j]+=(a.mat[i][k]*b.mat[k][j])%mod;
        ret.mat[i][j]=ret.mat[i][j]%mod;
    }
    return ret;
}

void calc(int k) {
    rep(i,0,n-1) rep(j,0,n-1) ansM.mat[i][j]=(i==j);
    while (k) {
        if (k&1) ansM=multi(initM,ansM);
        initM=multi(initM,initM);
        k>>=1;
    }

}

int main(int argc, char const *argv[]) {
#ifdef LOCAL
    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);
    // freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);
#endif
    int t;
    SI(t);
    while(t--) {
        SII(n,k);
        rep(i,0,n-1) rep(j,0,n-1) cin >> initM.mat[i][j];
        if (k>=2) calc(k);
        int ans=0;
        rep(i,0,n-1) ans=(ans+ansM.mat[i][i])%mod;
        cout << ans << endl;
    }
    return 0;
}
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