标签：long target code href log typedef cte ati comment
题意： 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);
给出k和mod，求f(k)。
总结： 1、特别注意，矩阵相乘不满足交换律，即a*b != b*a。 2、感觉推方程有点困难。 3、矩阵初始化注意。
f(x-10) 0 0 0 0 0 0 0 0 0 ( first矩阵 )
f(x-9) 0 0 0 0 0 0 0 0 0
f(x-8) 0 0 0 0 0 0 0 0 0
f(x-7) 0 0 0 0 0 0 0 0 0
f(x-6) 0 0 0 0 0 0 0 0 0
f(x-5) 0 0 0 0 0 0 0 0 0
f(x-4) 0 0 0 0 0 0 0 0 0
f(x-3) 0 0 0 0 0 0 0 0 0
f(x-2) 0 0 0 0 0 0 0 0 0
f(x-1) 0 0 0 0 0 0 0 0 0
*
0 1 0 0 0 0 0 0 0 0 ( temp矩阵 )
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
a9 8 7 6 5 4 3 2 1 0 (这一行是ai)
=
f(x-9) 0 0 0 0 0 0 0 0 0
f(x-8) 0 0 0 0 0 0 0 0 0
f(x-7) 0 0 0 0 0 0 0 0 0
f(x-6) 0 0 0 0 0 0 0 0 0
f(x-5) 0 0 0 0 0 0 0 0 0
f(x-4) 0 0 0 0 0 0 0 0 0
f(x-3) 0 0 0 0 0 0 0 0 0
f(x-2) 0 0 0 0 0 0 0 0 0
f(x-1) 0 0 0 0 0 0 0 0 0
f(x) 0 0 0 0 0 0 0 0 0
#include<iostream> #include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<queue> #include<stack> #include<map> #include<bitset> #include<vector> #include<set> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define F(i,a,b) for (int i=a;i<b;i++) #define FF(i,a,b) for (int i=a;i<=b;i++) #define mes(a,b) memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f typedef long long ll; const int N = 1e5+10, Maxn = 15; ll k,m=100; struct Mat { ll mat[Maxn][Maxn]; Mat() { mes(mat,0); F(i,0,Maxn) mat[i][i]=1; } } E; Mat first, temp; Mat operator * (Mat a, Mat b) { Mat ans; memset(ans.mat, 0, sizeof(ans.mat)); for (int i = 0; i < Maxn; i++) for (int j = 0; j < Maxn; j++) for (int k = 0; k < Maxn; k++) ans.mat[i][j] = (ans.mat[i][j] + a.mat[i][k] * b.mat[k][j]) % m; return ans; } Mat operator ^ (Mat a,ll x) { Mat p = E, q = a; while (x) { if(x&1) p = p*q; x>>=1; q = q*q; } return p; } int main() { mes(first.mat, 0); F(i,0,10) first.mat[i][0]=i; while(~scanf("%lld%lld", &k, &m)) { mes(temp.mat, 0); FF(i,0,8) temp.mat[i][i+1]=1; F(i,0,10) { scanf("%lld", &temp.mat[9][9-i]); } if(k<10) { printf("%lld\n", k); continue; } Mat ans= (temp^(k-9))*first; //注意，这里顺序不要反了 printf("%lld\n", ans.mat[9][0]); } return 0; }
矩阵模板
const int MOD , Maxn; //MOD作余，Maxn为矩阵范围 struct Mat { ll mat[Maxn][Maxn]; //开的long long Mat() { mes(mat,0); F(i,0,Maxn) mat[i][i]=1; //对角线初始化为1，其它为0 } } E; //单位矩阵 Mat first, temp; Mat operator * (Mat a, Mat b) //重载 * ，特别注意，矩阵相乘不满足交换律，即a*b != b*a { Mat ans; memset(ans.mat, 0, sizeof(ans.mat)); for (int i = 0; i < Maxn; i++) for (int j = 0; j < Maxn; j++) for (int k = 0; k < Maxn; k++) ans.mat[i][j] = (ans.mat[i][j] + a.mat[i][k] * b.mat[k][j]) % MOD; return ans; } Mat operator ^ (Mat a,ll x) ////重载 ^ { Mat p = E, q = a; while (x) { if(x&1) p = p*q; x>>=1; q = q*q; } return p; } Mat mul(Mat a,Mat b) //函数 * { Mat ans; int i,j,k; for(i=0;i<Maxn;i++){ for(j=0;j<Maxn;j++) { ans.mat[i][j]=0; for(k=0;k<Maxn;k++){ ans.mat[i][j]+=a.mat[i][k]*b.mat[k][j]%m; ans.mat[i][j]%=m; } } } return ans; } Mat Pow_Mod(Mat s, int b){ //函数开方 Mat ans; for(int i=0; i<Maxn; ++i) ans.mat[i][i]=1; while(b){ if(b&1) ans=mul(ans, s); s= mul(s, s); b>>=1; } return ans; } void prMat(Mat a) //打印矩阵 { int i,j; for(i=0;i<Maxn;i++) { for(j=0;j<Maxn;j++) printf("%d ",a.mat[i][j]); puts(""); } puts(""); return ; }
标签：long target code href log typedef cte ati comment
原文地址：http://www.cnblogs.com/sbfhy/p/6295386.html
