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

最新推荐文章于 2024-08-17 03:23:51 发布

转载最新推荐文章于 2024-08-17 03:23:51 发布 · 59 阅读

本文介绍了一种使用矩阵快速幂解决特定递推问题的方法。通过构建特定矩阵并利用快速幂运算，可以高效地计算出给定递推序列的第k项对m取模的结果。适用于处理大规模数据输入的情况。

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

把问题转化为求矩阵的n-9次幂就行了；

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 int k,MOD;
 6 int a[10];
 7 int f[10];
 8 struct Matrix
 9 {
10     int m[10][10];
11 }matrix;
12 
13 Matrix Mul(Matrix a,Matrix b)
14 {
15     Matrix res;
16     int i,j,k;
17     for(i=0;i<10;i++)
18     {
19         for(j=0;j<10;j++)
20         {
21             res.m[i][j] = 0;
22             for(k=0;k<10;k++)
23                 res.m[i][j] = (res.m[i][j]+(a.m[i][k]*b.m[k][j]))%MOD;
24         }
25     }
26     return res;
27 }
28 
29 Matrix fastm(Matrix a,int b)
30 {
31     Matrix res;
32     memset(res.m,0,sizeof(res.m));
33         for(int i=0;i<10;i++)
34             res.m[i][i] = 1;
35     while(b)
36     {
37         if(b&1)
38             res = Mul(res,a);
39         a = Mul(a,a);
40         b >>= 1;
41     }
42     return res;
43 }
44 void init()
45 {
46     for(int i=0;i<=9;i++)
47     {
48         f[i]=i;
49     }
50 }
51 int main()
52 {
53     init();
54     while(scanf("%d%d",&k,&MOD)==2)
55     {
56         for(int i=0;i<=9;i++)
57         {
58             scanf("%d",&a[i]);
59         }
60         if(k<10)
61         {
62             printf("%d\n",k%MOD);
63             continue;
64         }
65         
66         memset(matrix.m,0,sizeof(matrix.m));
67         for(int i=0;i<=9;i++)
68           matrix.m[0][i]=a[i];
69         for(int i=1;i<=9;i++)
70             matrix.m[i][i-1] = 1;
71         Matrix ans=fastm(matrix,k-9);
72         Matrix cnt;
73         for(int i=0;i<10;i++)
74         {
75             cnt.m[i][0]=f[9-i];
76         }
77         Matrix p=Mul(ans,cnt);
78         printf("%d\n",p.m[0][0]%MOD);
79     }
80     return 0;
81 }