HDU.1575 Tr A ( 矩阵快速幂)

HDU.1575 Tr A ( 矩阵快速幂)

点我挑战题目

题意分析

直接求矩阵A^K的结果，然后计算正对角线，即左上到右下对角线的和，结果模9973后输出即可。

由于此题矩阵直接给出的，题目比较裸。

代码总览

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <sstream>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#define INF 0x3f3f3f3f#define nmax 200#define MEM(x) memset(x,0,sizeof(x))using namespace std;const int Dmax = 11;int N;int MOD;typedef struct{    int matrix[Dmax][Dmax];    void init()//初始化为单位矩阵    {        memset(matrix,0,sizeof(matrix));        for(int i = 0; i<Dmax;++i) matrix[i][i] = 1;    }}MAT;MAT ADD(MAT a, MAT b){    for(int i = 0; i<N;++i){        for(int j = 0;j<N;++j){            a.matrix[i][j] +=b.matrix[i][j];            a.matrix[i][j] %= MOD;        }    }    return a;}MAT MUL(MAT a, MAT b){    MAT ans;    for(int i = 0; i<N;++i){        for(int j = 0; j<N;++j){            ans.matrix[i][j] = 0;            for(int k = 0; k<N;++k){                ans.matrix[i][j] += ( (a.matrix[i][k]) % MOD * (b.matrix[k][j]) % MOD) % MOD;            }            ans.matrix[i][j] %= MOD;        }    }    return ans;}MAT POW(MAT a, int t){    MAT ans; ans.init();    while(t){        if(t&1) ans = MUL(ans,a);        t>>=1;        a = MUL(a,a);    }    return ans;}void OUT(MAT a){    for(int i = 0; i<N;++i){        for(int j =  0; j<N;++j){            printf("%5d",a.matrix[i][j]);        }        printf("\n");    }}void IN(MAT & a){    for(int i = 0; i<N;++i){        for(int j = 0; j<N;++j){            scanf("%d",&a.matrix[i][j]);        }    }}void CAL(MAT a){    long long ans = 0;    for(int i = 0; i<N;++i) ans+=a.matrix[i][i];    ans %= MOD;    printf("%lld\n",ans);}int main(){    //freopen("in.txt","r",stdin);    int T;    scanf("%d",&T);    while(T--){        MOD = 9973;        int k;MAT m;        scanf("%d %d",&N,&k);        IN(m);        m = POW(m,k);        CAL(m);    }    return 0;}