hdu1757 A Simple Math Problem （矩阵快速幂）

最新推荐文章于 2019-05-14 15:56:22 发布

原创最新推荐文章于 2019-05-14 15:56:22 发布 · 569 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一个递归数学问题的解决方法，通过矩阵快速幂算法高效计算特定递推数列的第k项并对m取模。该问题涉及递推公式定义及边界条件，并通过输入示例展示了具体的求解过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3762 Accepted Submission(s): 2269

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

Author

linle

Source

2007省赛集训队练习赛（6）_linle专场

Recommend

lcy | We have carefully selected several similar problems for you: 1588 3117 2276 2256 2254

代码：

#include<cstdio>
using namespace std;

int m;

void multi(int (*x)[10],int (*y)[10])
{
  int i,j,k,c[10][10]={0};
  for(i=0;i<10;i++)
    for(j=0;j<10;j++)
      for(k=0;k<10;k++)
        c[i][j]+=x[i][k]*y[k][j];
  for(i=0;i<10;i++)
    for(j=0;j<10;j++)
      x[i][j]=c[i][j]%m; 
}

int main()
{
  //freopen("1.in","r",stdin);
  
  int k,i,j,a[10][10],b[10][10];
  while(scanf("%d%d",&k,&m)==2)
    {
      for(i=0;i<10;i++)scanf("%d",&a[0][i]);
      if(k<10){printf("%d\n",k%m);continue;}
      
      for(i=1;i<10;i++)
        for(j=0;j<10;j++)
          a[i][j]=(i-j==1);
      for(i=0;i<10;i++)
        for(j=0;j<10;j++)
          b[i][j]=(i==j);
          
	  for(k-=9;k;multi(a,a),k>>=1)if(k&1)multi(b,a);
      for(j=i=0;i<10;i++)j+=b[0][i]*(9-i);
      printf("%d\n",j%m);
	}
  return 0;
}