hdoj 1757 A Simple Math Problem(转化为矩阵+矩阵乘法)

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【题目大意】：If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);

And ai(0<=i<=9) can only be 0 or 1 .

先给出ai,和x,m，求f(x)%m

【解题思路】：

根据题目意思构造矩阵，跑矩阵乘法即可

【代码】：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>#include <cmath>#include <string>#include <cctype>#include <map>#include <iomanip>                   using namespace std;                   #define eps 1e-8#define pi acos(-1.0)#define inf 1<<30#define linf 1LL<<60#define pb push_back#define lc(x) (x << 1)#define rc(x) (x << 1 | 1)#define lowbit(x) (x & (-x))#define ll long long  struct Mart{      int mat[105][105];  };  Mart mm;int n;  Mart y;  int k,m,i;  int sum;        Mart mart_multiply(Mart a,Mart b){      Mart c;      for(int i=1; i<=10; i++)          for(int j=1; j<=10; j++){              c.mat[i][j]=0;              for(int k=1; k<=10; k++)                  c.mat[i][j]=(c.mat[i][j]%m+a.mat[i][k]*b.mat[k][j]%m)%m;          }      return c;  }    void copy(Mart p){      for (int i=1; i<=10; i++)          for (int j=1; j<=10; j++)              y.mat[i][j]=p.mat[i][j];  }    void solve_mart_power(int np){      Mart p,q;      for(int i=1; i<=10; i++)          for(int j=1; j<=10; j++){              p.mat[i][j]=mm.mat[i][j];              if(i==j)  q.mat[i][j]=1.0;              else  q.mat[i][j]=0.0;          }      if(np==0) copy(q);      else {          while(np!=1){              if(np&1){                  np--;                  q=mart_multiply(p,q);              }              else{                  np=np/2;                  p=mart_multiply(p,p);              }          }          p=mart_multiply(p,q);          copy(p);      }      return ;  }  int main()  {    for (int i=1; i<=10; i++){mm.mat[i+1][i]=1; mm.mat[1][i]=1;}              while (~scanf("%d%d",&k,&m))  {          for (int i=1; i<=10; i++)              scanf("%d", &mm.mat[1][i]);          if (k<10) {              printf("%d/n",k%m);          }        else {              solve_mart_power(k-9);              sum=0;              for (int i=1; i<=10; i++)                  sum+=y.mat[1][i]*(10-i);              printf("%d\n", sum % m);        }    }    return 0;  }