HDU 1757 矩陣相乘，快速冪模板題

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HDU 1757

題意： 　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); 

　　　　  給出k和mod，求f(k)。

總結： 1、特別注意，矩陣相乘不滿足交換律，即a*b != b*a。  2、感覺推方程有點困難。 3、矩陣初始化注意。

f(x-10)   0 0 0 0 0 0 0 0 0        ( first矩陣 )      
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矩陣 )
  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  (這一行是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 0x3f3f3f3ftypedef 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，這裡順序不要反了        printf("%lld\n", ans.mat[9][0]);    }    return 0;}

View Code

 

矩陣模板

const int MOD , Maxn;  //MOD作餘，Maxn為矩陣範圍  struct Mat{    ll mat[Maxn][Maxn];     //開的long long    Mat() {        mes(mat,0);        F(i,0,Maxn) mat[i][i]=1;    //對角線初始化為1，其它為0    }} E;   //單位矩陣Mat first, temp;Mat operator * (Mat a, Mat b)   //重載 * ，特別注意，矩陣相乘不滿足交換律，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)    ////重載 ^{    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)    //函數 *{    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){   //函數開方    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)   //列印矩陣{    int i,j;    for(i=0;i<Maxn;i++)    {        for(j=0;j<Maxn;j++)            printf("%d ",a.mat[i][j]);        puts("");    }    puts("");    return ;}

 

HDU 1757 矩陣相乘，快速冪模板題