快速幂
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inline int power(int a,int b,int p)
{
long long ans=1;
while(b)
{
if(b&1)ans=ans*a%p;
a=a*a%p;
b>>=1;
}
return ans;
}
矩阵快速幂
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struct Matrix
{
int n,m;
int a[maxn][maxn];
Matrix(){memset(a,0,sizeof a);}
};
Matrix operator *(Matrix a,Matrix b)
{
Matrix tmp;
tmp.n=a.n,tmp.m=b.m;
for(int i=1;i<=a.n;++i)
{
for(int j=1;j<=b.m;++j)
{
for(int k=1;k<=a.m;++k)
{
tmp.a[i][j]=(tmp.a[i][j]+a.a[i][k]*b.a[k][j])%MOD;
}
}
}
return tmp;
}
Matrix base(int x)
{
Matrix tmp;
tmp.n=tmp.m=x;
for(int i=1;i<=x;++i)tmp.a[i][i]=1;
return tmp;
}
Matrix power(Matrix a,int b)
{
Matrix ans=base(a.n);
while(b)
{
if(b&1)ans=ans*a;
a=a*a;
b>>=1;
}
return ans;
}