HDU 1575 Tr A

http://acm.hdu.edu.cn/showproblem.php?pid=1575

題意 ： 裸裸的矩陣相乘

思路 ： 因為K比較大，所以我使用二進位最佳化，最後只要把主對角線上的數字加起來最後取下模即可。

//Danceonly#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef long long LL;const double INF = 100000007;const double eps = 1e-9;const int maxn = 15;const int mod = 9973;#define MIN(a,b) (a > b ? b : a)#define MAX(a,b) (a > b ? a : b)int maze[40][maxn][maxn];int ans[maxn][maxn],temp[maxn][maxn];bool tt[40];int N,K;void Fu(){      for (int i=1;i<=N;i++)            for (int j=1;j<=N;j++)                  ans[i][j] = temp[i][j];}void init(int a){      for (int i=1;i<=N;i++)            for (int j=1;j<=N;j++)            {                  int sum = 0;                  for (int k=1;k<=N;k++)                        sum += (maze[a][i][k] * maze[a][k][j]) % mod;                  maze[a + 1][i][j] = sum % mod;            }}void Add(int n){      for (int i=1;i<=N;i++)            for (int j=1;j<=N;j++)            {                  int sum = 0;                  for (int k=1;k<=N;k++)                        sum += (ans[i][k]  * maze[n][k][j]) % mod;                  temp[i][j] = sum % mod;            }      Fu();}void show(){      for (int k=1;k<=10;k++)      {            printf("k = %d\n",k);            for (int i=1;i<=N;i++)            {                  for (int j=1;j<=N;j++)                        printf("%d ",maze[k][i][j]);                  printf("\n");            }            printf("\n");      }}int main(){      int T;      scanf("%d",&T);      while (T--)      {            scanf("%d%d",&N,&K);            for (int i=1;i<=N;i++)                  for (int j=1;j<=N;j++)                  {                        scanf("%d",&maze[1][i][j]);                        ans[i][j] = maze[1][i][j];                  }            for (int i=1;i<=32;i++)                  init(i);            //show();            int e = 1,cnt = 0;            while (K >= e)            {                  e *= 2;                  cnt ++;            }            memset(tt,0,sizeof(tt));            for (int i=0;K;i++)            {                  if (K >= e)                  {                        K -= e;                        tt[cnt - i] = 1;                  }                  else                        tt[cnt - i] = 0;                  e /= 2;            }            for (int i=1;i<=cnt;i++)                  if (tt[i]) Add(i+1);            int sum = 0;            for (int i=1;i<=N;i++)                  sum += ans[i][i];            printf("%d\n",sum % mod);      }      return 0;}