10 9999 1 1 1 1 1 1 1 1 1 1 20 500 1 0 1 0 1 0 1 0 1 0
45 104 CODE:#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<string> #include<algorithm> #include<cstdlib> #include<set> #include<queue> #include<stack> #include<vector> #include<map> #define N 100010 #define lson l,mid,idx<<1 #define rson mid+1,r,idx<<1|1 #define lc idx<<1 #define rc idx<<1|1 const double EPS = 1e-11; const double PI = acos ( -1.0 ); const double E = 2.718281828; typedef long long ll; const int INF = 1000010; using namespace std; typedef vector<int>vec; typedef vector<vec>mat; int m,fac[11]; ll k; void init() { for(int i=0;i<10;i++) fac[i]=i; } mat mul(mat &A,mat &B,int Mod) { mat C(A.size(),vec(B[0].size())); for(int i=0; i<A.size(); i++) { for(int k=0; k<B.size(); k++) { for(int j=0; j<B[0].size(); j++) { C[i][j]=(C[i][j]+A[i][k]*B[k][j])%Mod; } } } return C; } mat pow(mat A,ll n,int Mod) { mat B(A.size(),vec(A.size())); for(int i=0; i<A.size(); i++) { B[i][i]=1; } while(n>0) { if(n&1) B=mul(B,A,Mod); A=mul(A,A,Mod); n>>=1; } return B; } int main() { init(); while(cin>>k>>m) { if(k<10) { printf("%d\n",fac[k]%m); continue; } mat A(10,vec(10)); for(int i=0; i<10; i++) { scanf("%d",&A[0][i]); } for(int i=1;i<10;i++) { for(int j=0;j<10;j++) { if(i==j+1) A[i][j]=1; else A[i][j]=0; } } A=pow(A,k-9,m); int ans=0; for(int i=0;i<10;i++) { ans+=A[0][i]*fac[9-i]; if(ans>=m) ans%=m; } cout<<ans<<endl; } return 0; }
HDU 1757 A Simple Math Problem(矩阵快速幂)
原文地址:http://blog.csdn.net/acm_baihuzi/article/details/43229193