hdu1757 A Simple Math Problem【矩阵快速幂】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1757

题意：题目有一个表达式f(x)，让你求他的第k项，即f(k)%m

解析：应该就是一个矩阵快速幂了，构造一个矩阵，草稿纸上算一下不难得出这个矩阵

[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

0,0,0,0,0,0,0,0,1,0

0,0,0,0,0,0,0,0,0,1

a9,a8,a7,a6,a5,a4,a3,a2,a1,a0]

#include <bits/stdc++.h>
using namespace std;
int k,mod;
struct matrix
{
long long a[15][15];
int n,m;
matrix() {}
matrix(int _n,int _m)
{
n = _n,m = _m;
memset(a,0,sizeof(a));
}
matrix operator * (const matrix &b)const
{
matrix res(n,b.m);
for(int i=0;i<n;i++)
{
for(int j=0;j<b.m;j++)
{
for(int k=0;k<m;k++)
res.a[i][j] = (res.a[i][j]+a[i][k]*b.a[k][j])%mod;
}
}
return res;
}
void print()
{
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
printf("%I64d ",a[i][j]);
puts("");
}
}
};
matrix qpow(matrix x,int n)
{
matrix res(x.n,x.m);
for(int i=0;i<res.n;i++)
res.a[i][i] = 1;
while(n>0)
{
if(n&1)
res = res*x;
x = x*x;
n >>= 1;
}
return res;
}
int main(void)
{
while(~scanf("%d %d",&k,&mod))
{
matrix op(10,10);
matrix ans(10,1);
for(int i=0;i<10;i++)
ans.a[i][0] = i;
for(int i=0;i<9;i++)
op.a[i][i+1] = 1;
for(int i=9;i>=0;i--)
scanf("%I64d",&op.a[9][i]);
//op.print();
op = qpow(op,k);
//op.print();
ans = op*ans;
//ans.print();
printf("%I64d\n",ans.a[0][0]);
}
return 0;
}