首页 > 代码库 > hdu 1757 A Simple Math Problem（矩阵快速幂）

hdu 1757 A Simple Math Problem（矩阵快速幂）

2024-07-23 02:03:26   227人阅读

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2780    Accepted Submission(s): 1649


Problem Description
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.
 

Input
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.
 

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

Sample Input
10 9999
1 1 1 1 1 1 1 1 1 1
20 500
1 0 1 0 1 0 1 0 1 0
 

Sample Output
45
104
 
图解：



这个图来自：http://www.cnblogs.com/wally/archive/2013/03/01/2938305.html

#include"iostream"
#include"stdio.h"
#include"string.h"
#include"algorithm"
#include"queue"
#include"vector"
using namespace std;
#define N 10
#define LL __int64
int M,f[N];
struct Mat
{
    LL mat[N][N];
};
Mat operator *(Mat a,Mat b)
{
    int i,j,k;
    Mat c;
    for(i=0;i<N;i++)
    {
        for(j=0;j<N;j++)
        {
            c.mat[i][j]=0;
            for(k=0;k<N;k++)
            {
                c.mat[i][j]+=(a.mat[i][k]*b.mat[k][j])%M;
            }
        }
    }
    return c;
}
int fun(Mat &a,int k)
{
    int i;
    Mat ans;
    memset(ans.mat,0,sizeof(ans.mat));
    k-=9;
    for(i=0;i<N;i++)
        ans.mat[i][i]=1;
    for(i=1;i<N;i++)
        a.mat[i][i-1]=1;
    while(k)
    {
        if(k&1)
            ans=ans*a;
        k>>=1;
        a=a*a;
    }
    LL s=0;
    for(i=0;i<N;i++)
    {
        s+=ans.mat[0][i]*f[N-i-1]%M;
        s%=M;
    }
    return s;
}
int main()
{
    int k,i;
    Mat a;
    while(scanf("%d%d",&k,&M)!=-1)
    {
        memset(a.mat,0,sizeof(a.mat));
        for(i=0;i<N;i++)
        {
            scanf("%I64d",&a.mat[0][i]);
            f[i]=i;
        }
        if(k<N)
            printf("%d\n",i%M);
        else
            printf("%d\n",fun(a,k));
    }
    return 0;
}



hdu 1757 A Simple Math Problem（矩阵快速幂）


联系
我们