【HDU 1757】 A Simple Math Problem

饶文津

发布于 2020-05-31 23:40:16

4090

发布于 2020-05-31 23:40:16

文章被收录于专栏：饶文津的专栏

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

题意：按f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10) (x>=10) ； f(x) = x（x<10）来计算f(x)%m的值。

分析：这题要用递推，并且k值很大，所以需要用矩阵快速幂。

构造的矩阵是：

a0 a1 a2 a3 a4 a5 a6 a7 a8 a9 1 1 1 1 1 1 1 1 1	a0	a1	a2	a3	a4	a5	a6	a7	a8
a0	a1	a2	a3	a4	a5	a6	a7	a8	a9
1
	1
		1
			1
				1
					1
						1
							1
								1
f(x-1)
f(x-2)
f(x-3)
f(x-4)
f(x-5)
f(x-6)
f(x-7)
f(x-8)
f(x-9)
f(x-10)
f(x)
f(x-1)
f(x-2)
f(x-3)
f(x-4)
f(x-5)
f(x-6)
f(x-7)
f(x-8)
f(x-9)

写个结构类型代表矩阵，以及矩阵的相乘的函数和矩阵快速幂的函数，注意一下初始化。

#include<stdio.h>
#include<string.h>
int n,k,m;
struct matrix
{
    int a[15][15];
    int row,col;
    void init(int r,int c){
        memset(a,0,sizeof(a));
        row=r;col=c;
    }
} big,f,u;
matrix mul(matrix a,matrix b)
{
    matrix c;
    c.init(a.row,b.col);
    for(int i=0; i<a.row; i++)
        for(int j=0; j<b.col; j++)
        {
            for(int k=0; k<a.col; k++)
                c.a[i][j]+=(a.a[i][k]*b.a[k][j])%m;
            c.a[i][j]%=m;
        }
    return c;
}
void init()
{
    big.init(10,10);
    f.init(10,1);
    for(int i=1; i<10; i++)
        big.a[i][i-1]=1;
    for(int i=0; i<10; i++)
        f.a[i][0]=9-i;
}
matrix qpow(matrix a,int k)
{
    matrix ans;
    ans.init(a.row,a.col);
    for(int i=0;i<a.row;i++)
        ans.a[i][i]=1;
    while(k)
    {
        if(k&1)ans=mul(ans,a);
        a=mul(a,a);
        k>>=1;
    }
    return ans;
}
int main()
{
    init();
    while(~scanf("%d%d",&k,&m))
    {
        for(int i=0; i<10; i++)
            scanf("%d",&big.a[0][i]);
        u=mul(qpow(big,k-9),f);
        printf("%d\n",u.a[0][0]%m);
    }
    return 0;
}

本文参与腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。

原始发表：2016-02-06 ，如有侵权请联系 cloudcommunity@tencent.com 删除

case