hdu 1757 A Simple Math Problem（矩阵快速幂乘法）

[align=left]Problem Description[/align]

Lele now is thinking about a simple function f(x).

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.

[align=left]Input[/align]

The problem contains mutiple test cases.Please process to the end of file.
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.

[align=left]Output[/align]

For each case, output f(k) % m in one line.

[align=left]Sample Input[/align]

10 9999
1 1 1 1 1 1 1 1 1 1
20 500
1 0 1 0 1 0 1 0 1 0

[align=left]Sample Output[/align]

45
104

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int k,MOD;
int a[10];
int f[10];
struct Matrix
{
int m[10][10];
}matrix;

Matrix Mul(Matrix a,Matrix b)
{
Matrix res;
int i,j,k;
for(i=0;i<10;i++)
{
for(j=0;j<10;j++)
{
res.m[i][j] = 0;
for(k=0;k<10;k++)
res.m[i][j] = (res.m[i][j]+(a.m[i][k]*b.m[k][j]))%MOD;
}
}
return res;
}

Matrix fastm(Matrix a,int b)
{
Matrix res;
memset(res.m,0,sizeof(res.m));
for(int i=0;i<10;i++)
res.m[i][i] = 1;
while(b)
{
if(b&1)
res = Mul(res,a);
a = Mul(a,a);
b >>= 1;
}
return res;
}
void init()
{
for(int i=0;i<=9;i++)
{
f[i]=i;
}
}
int main()
{
init();
while(scanf("%d%d",&k,&MOD)==2)
{
for(int i=0;i<=9;i++)
{
scanf("%d",&a[i]);
}
if(k<10)
{
printf("%d\n",k%MOD);
continue;
}

memset(matrix.m,0,sizeof(matrix.m));
for(int i=0;i<=9;i++)
matrix.m[0][i]=a[i];
for(int i=1;i<=9;i++)
matrix.m[i][i-1] = 1;
Matrix ans=fastm(matrix,k-9);
Matrix cnt;
for(int i=0;i<10;i++)
{
cnt.m[i][0]=f[9-i];
}
Matrix p=Mul(ans,cnt);
printf("%d\n",p.m[0][0]%MOD);
}
return 0;
}

View Code