2018-5-24

簡單的矩陣快速冪問題，重點是如何找到對應關系。

f(10) a0 a1 a2 a3 a4 a5 a6 a7 a8 a9 f(9) f(9) 1 0 0 0 0 0 0 0 0 0 f(8) f(8) 0 1 0 0 0 0 0 0 0 0 f(7) f(7) 0 0 1 0 0 0 0 0 0 0 f(6) f(6) = 0 0 0 1 0 0 0 0 0 0 * f(5) f(5) 0 0 0 0 1 0 0 0 0 0 f(4) f(4) 0 0 0 0 0 1 0 0 0 0 f(3) f(3) 0 0 0 0 0 0 1 0 0 0 f(2) f(2) 0 0 0 0 0 0 0 1 0 0 f(1) f(1) 0 0 0 0 0 0 0 0 1 0 f(0) #include<iostream> #include<cstring> using namespace std;const int N = 10; int x[N+1][N+1]; int k,m;void cal(int p[N+1][N+1],int q[N+1][N+1],int r[N+1][N+1]){for (int i=0;i<N;i++){for (int j=0;j<N;j++){for (int k=0;k<N;k++){r[i][j]+=p[i][k]*q[k][j];r[i][j]%=m;}}} }void fun(int b){int res[N+1][N+1],base[N+1][N+1],tmp[N+1][N+1];memset(res,0,sizeof(res));for (int i=0;i<N;i++) res[i][i]=1;memcpy(base,x,sizeof(base));while (b){if (b&1==1){memset(tmp,0,sizeof(tmp));cal(res,base,tmp);memcpy(res,tmp,sizeof(res));}memset(tmp,0,sizeof(tmp));cal(base,base,tmp);memcpy(base,tmp,sizeof(base));b>>=1;}int result=0;for (int i=0;i<N;i++){result+=res[0][i]*(N-i-1);result%=m;}cout<<result<<endl; }int main(){while (cin>>k>>m){memset(x,0,sizeof(x));for (int i=0;i<N;i++){cin>>x[0][i];}for (int i=1;i<N;i++){x[i][i-1]=1;}if (k<10) cout<<k%m<<endl;else{fun(k-9);}}return 0; }/* 20 500 1 0 1 0 1 0 1 0 1 0 */

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