A Simple Math Problem
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4334 Accepted Submission(s): 2605
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);
And ai(0<=i<=9) can only be 0 or 1 .
Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.
104
矩陣快速冪。
代碼:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int MOD;
struct mat
{
int a[10][10];
};
mat mat_mul(mat x,mat y)
{
mat res;
memset(res.a,0,sizeof(res.a));
for(int i=0; i<10; i++)
{
for(int j=0; j<10; j++)
{
for(int k=0; k<10; k++)
{
if(x.a[i][k]&&y.a[k][j])
{
res.a[i][j]=(res.a[i][j]+x.a[i][k]*y.a[k][j])%MOD;
}
}
}
}
return res;
}
mat mat_fast(mat c,int N)
{
mat res;
memset(res.a,0,sizeof(res.a));
for(int i=0; i<10; i++)
{
res.a[i][i]=1;
}
while(N!=0)
{
if(N&1)
{
res=mat_mul(res,c);
}
c=mat_mul(c,c);
N=N>>1;
}
//printf("%I64d\n",res.a[0][1]);
return res;
}
int main()
{
int k;
while(~scanf("%d%d",&k,&MOD))
{
if(k<10)
{
printf("%d\n",k);
continue;
}
int x;mat c,res;
memset(c.a,0,sizeof(c.a));
memset(res.a,0,sizeof(res.a));
for(int i=0; i<10; i++)
{
scanf("%d",&x);
c.a[0][i]=x;
res.a[i][0]=9-i;
}
for(int i=1; i<10; i++)
{
c.a[i][i-1]=1;
}
mat ans=mat_mul(mat_fast(c,k-9),res);
printf("%d\n",ans.a[0][0]%MOD);
}
}