hdu 1757 A Simple Math Problem（矩阵快速幂）

最新推荐文章于 2019-07-02 13:43:53 发布

KinFungYu

最新推荐文章于 2019-07-02 13:43:53 发布

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分类专栏：矩阵 hdu 文章标签：矩阵快速幂

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本文介绍了一种通过构造矩阵的方法解决特定递归数列问题的算法，并提供了完整的AC代码实现。该方法适用于当输入参数较大时，高效计算数列中指定项的值并对结果取模。

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A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2051 Accepted Submission(s): 1178

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

思路：构造矩阵

|fn| |a0 a1 a2....a9| |fn-1|

|fn-1| |1 0 0.... 0 | |fn-2|

|fn-2| = |0 1 0.... 0| * |fn-3|

...... ..... .....

|fn-9| |0 0 .... 1 0| |fn-10|

AC代码：

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <queue>
#include <stack>
#include <ctime>
#include <vector>
#include <algorithm>
#define ll __int64
#define L(rt) (rt<<1)
#define R(rt)  (rt<<1|1)

using namespace std;

const int INF = 2000000000;
const int maxn = 10;
struct Mat
{
    int mat[maxn][maxn];
    void zero()
    {
        memset(mat, 0, sizeof(mat));
    }
    void unit()
    {
        zero();
        for(int i = 0; i < maxn; i++) mat[i][i] = 1;
    }
}A, T;
int a[10];
int k, m;
Mat operator *(const Mat &a, const Mat &b)
{
    Mat tmp;
    for(int i = 0; i < maxn; i++)
        for(int j = 0; j < maxn; j++)
        {
            int sum = 0;
            for(int k = 0; k < maxn; k++)
                sum += a.mat[i][k] * b.mat[k][j];
            tmp.mat[i][j] = sum % m;
        }
    return tmp;
}
Mat operator ^ (Mat x, int n){
    Mat tmp;
    tmp.unit();
    while(n)
    {
        if(n & 1) tmp = tmp * x;
        x = x * x;
        n >>= 1;
    }
    return tmp;
}
void init(){
    A.zero();
    for(int i = 0; i < maxn; i++)
    A.mat[i][0] = maxn - 1 - i;
    T.zero();
    for(int i = 0; i < maxn; i++)
    T.mat[0][i] = a[i];
    for(int i = 1; i < maxn; i++)
    T.mat[i][i - 1] = 1;
}
int main()
{
    while(~scanf("%d%d", &k, &m))
    {
        for(int i = 0; i < maxn; i++)
        scanf("%d", &a[i]);
        if(k < 10)
        {
            printf("%d\n", k);
            continue;
        }
        init();
        Mat ans = (T ^ (k - 9)) * A;
        printf("%d\n", ans.mat[0][0]);
    }
    return 0;
}