Recurrences
阿新 • • 發佈:2018-12-12
連結
模板
struct Matrix
{
ll n, m, a[30][30];
ll* operator[](ll x){return a[x];}
Matrix(ll x, ll y)
{
n=x, m=y;
cl(a);
}
void show()
{
ll i, j;
printf("Matrix %lld*%lld:\n",n,m);
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)printf("%lld ",a[i][j]);
putchar(10);
}
}
};
Matrix operator* (Matrix a, Matrix b)
{
Matrix t(a.n,b.m);
ll i, j, k;
for(i=1;i<=t.n;i++)for(j=1;j<=t.m;j++)for(k=1;k<=a.m;k++)t[i][j]=(t[i][j]+a[i][k]*b[k][j])%mod;
return t;
}
Matrix operator+(Matrix a, Matrix b)
{
Matrix t(a.n,a.m);
ll i, j;
for(i=1;i<=t.n;i++)for(j=1;j<=t.m;j++)t[i][j]=(a[ i][j]+b[i][j])%mod;
return t;
}
Matrix operator-(Matrix a, Matrix b)
{
Matrix t(a.n,a.m);
ll i, j;
for(i=1;i<=t.n;i++)for(j=1;j<=t.m;j++)t[i][j]=(a[i][j]-b[i][j]+mod)%mod;
return t;
}
Matrix fastpow(Matrix a, ll b)
{
ll i, j;
Matrix t(a.n,a.m), ans(a.n,a.m);
for(i=1;i<=a.n;i++)for(j= 1;j<=a.m;j++)t[i][j]=a[i][j];
for(i=1;i<=a.n;i++)ans[i][i]=1;
for(;b;b>>=1,t=t*t)if(b&1)ans=ans*t;
return ans;
}
宣告矩陣的時候必須要宣告行和列:Matrix m(a,b)
題解
矩陣快速冪模板題,一定要搞清楚行和列…我竟然調了一個多小時
程式碼
//矩陣
#include <bits/stdc++.h>
#define ll long long
#define cl(x) memset(x,0,sizeof(x))
using namespace std;
ll mod;
struct Matrix
{
ll n, m, a[30][30];
ll* operator[](ll x){return a[x];}
Matrix(ll x, ll y)
{
n=x, m=y;
cl(a);
}
void show()
{
ll i, j;
printf("Matrix %lld*%lld:\n",n,m);
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)printf("%lld ",a[i][j]);
putchar(10);
}
}
};
Matrix operator*(Matrix a, Matrix b)
{
Matrix t(a.n,b.m);
ll i, j, k;
for(i=1;i<=t.n;i++)for(j=1;j<=t.m;j++)for(k=1;k<=a.m;k++)t[i][j]=(t[i][j]+a[i][k]*b[k][j])%mod;
return t;
}
Matrix operator+(Matrix a, Matrix b)
{
Matrix t(a.n,a.m);
ll i, j;
for(i=1;i<=t.n;i++)for(j=1;j<=t.m;j++)t[i][j]=(a[i][j]+b[i][j])%mod;
return t;
}
Matrix operator-(Matrix a, Matrix b)
{
Matrix t(a.n,a.m);
ll i, j;
for(i=1;i<=t.n;i++)for(j=1;j<=t.m;j++)t[i][j]=(a[i][j]-b[i][j]+mod)%mod;
return t;
}
Matrix fastpow(Matrix a, ll b)
{
ll i, j;
Matrix t(a.n,a.m), ans(a.n,a.m);
for(i=1;i<=a.n;i++)for(j=1;j<=a.m;j++)t[i][j]=a[i][j];
for(i=1;i<=a.n;i++)ans[i][i]=1;
for(;b;b>>=1,t=t*t)if(b&1)ans=ans*t;
return ans;
}
int main()
{
ll d, n, a[20], i;
while(scanf("%lld%lld%lld",&d,&n,&mod),d)
{
Matrix f(d,1), trans(d,d);
for(i=1;i<d;i++)trans[i][i+1]=1;
for(i=1;i<=d;i++)scanf("%lld",&trans[d][d-i+1]);
for(i=1;i<=d;i++)scanf("%lld",&f[i][1]);
f=fastpow(trans,n-1)*f;
printf("%lld\n",f[1][1]);
}
return 0;
}