矩阵快速幂模板(1575)

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Tr A

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4060 Accepted Submission(s): 3031

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

/*------------------Header Files------------------*/#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <algorithm>#include <cstdlib>#include <ctype.h>#include <cmath>#include <stack>#include <queue>#include <deque>#include <map>#include <vector>#include <limits.h>using namespace std;/*------------------Definitions-------------------*/#define LL long long#define PI acos(-1.0)#define INF 0x3F3F3F3F#define MOD 9973#define MAX 500050#define lson rt<<1,l,m#define rson rt<<1|1,m+1,r/*---------------------Work-----------------------*/struct node{int num[11][11];int r,c;};node todo,sum;void init(int n){todo.r=todo.c=n;sum.r=sum.c=n;memset(sum.num,0,sizeof(sum.num));for(int i=1;i<=todo.r;i++){sum.num[i][i]=1; //初始化为单位矩阵for(int j=1;j<=todo.c;j++)scanf("%d",&todo.num[i][j]);}}node multi(node a,node b){node z;memset(z.num,0,sizeof(z.num));z.r=a.r,z.c=b.c;for(int i=1;i<=a.r;i++){for(int k=1;k<=a.c;k++){if(a.num[i][k]==0) continue;for(int j=1;j<=b.c;j++)z.num[i][j]=(z.num[i][j]+(a.num[i][k]*b.num[k][j])%MOD)%MOD;}}return z;}void matrix(int n){while(n){if(n&1) sum=multi(todo,sum);todo=multi(todo,todo);n>>=1;}int ans=0;for(int i=1;i<=sum.c;i++)ans=(ans+sum.num[i][i])%MOD;printf("%d\n",ans);}void work(){int T; cin>>T;while(T--){int n,k;scanf("%d%d",&n,&k);init(n);matrix(k);}}/*------------------Main Function------------------*/int main(){//freopen("test.txt","r",stdin);//freopen("cowtour.out","w",stdout);//freopen("cowtour.in","r",stdin);work();return 0;}

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