1. 程式人生 > >[矩陣快速冪]hdu1575 Tr A

[矩陣快速冪]hdu1575 Tr A

hdu1575 Tr A

題意:
求矩陣A的k次冪,主對角線之和%9973
思路:
裸的矩陣快速冪,具體見模板

程式碼:

/**************************************************************
    Problem: HDU_1575
    User: soundwave
    Language: C++
    Result: Accepted
    Time: 0ms
    Memory: 1576KB
****************************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <stdio.h>
#include <vector>

using namespace std;
typedef vector<int> Vint;
typedef vector<Vint> VVint;
typedef __int64 LL;
const int MOD = 9973;
//矩陣乘法
VVint calc(VVint &A, VVint &B){
    VVint C(A.size(),Vint(A.size()));
    for(int i=0; i<A.size(); i++)
    for(int j=0; j<B[0].size(); j++)
    for(int k=0; k<B.size(); k++)
        C[i][j] = (C[i][j] + (A[i][k]*B[k][j])) % MOD;
    return C;
}
//二分快速冪
VVint my_pow(VVint &A, int c){
    VVint B(A.size(),Vint(A.size()));
    for(int i=0; i<A.size(); i++)
        B[i][i] = 1;
    if(c==1) return A;
    while(c>0){
        if(c&1) B = calc(B,A);
        A = calc(A,A);
        c>>=1;
    }
    return B;
}
int main(){
    int t, n, k;
    scanf("%d", &t);
    while(t-->0){
        scanf("%d%d", &n, &k);
        VVint A(n,Vint(n));
        for(int i=0; i<n; i++)
        for(int j=0; j<n; j++)
            scanf("%d", &A[i][j]);
        A = my_pow(A,k);
        LL re=0;
        for(int i=0; i<n; i++)
            re += A[i][i];
        printf("%I64d\n", re%MOD);
    }
    return 0;
}