【HDU 1757 A Simple Math Problem】+ 矩阵

最新推荐文章于 2019-05-14 15:56:22 发布

楚江枫

最新推荐文章于 2019-05-14 15:56:22 发布

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分类专栏：矩阵快速幂文章标签： math

本文链接：https://blog.youkuaiyun.com/WYK1823376647/article/details/53031617

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本文介绍了一个简单的数学问题，该问题涉及递归函数计算及利用矩阵快速幂算法进行高效求解的方法。对于给定的参数，文章提供了一种计算特定递归函数值的有效途径，并附带AC代码实现。

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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): 4285 Accepted Submission(s): 2574

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

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);

|f(10) | |a0 a1 a2 …a8 a9| |f(9)|
| f(9) | | 1 0 0 … 0 0 | |f(8)|
| ….. | = | .. … … … | | .. |
| f(2) | | 0 0 0 … 0 0| |f(1)|
| f(1) |　　 | 0 0 0 … 1 0| |f(0)|

另A举证为10*10的举证，如上图。
可以推出：
(f(n),f(n-1),…,f(n-9))^(-1) = A^(n-9)*(f(9),f(8),…,f(0))^(-1)

矩阵～～现代～～～全忘光了～～

AC代码：

#include<bits/stdc++.h>
typedef long long LL;
LL K,mod;
struct node{
    int m[12][12];
}in,un;
void init(){ // 矩阵初始化
    memset(in.m,0,sizeof(in.m));
    memset(un.m,0,sizeof(un.m));
    for(int i = 1 ; i <= 9 ; i++)
        in.m[i][i - 1] = 1;
    for(int i = 0 ; i <= 9 ; i++) // 单位矩阵
        un.m[i][i] = 1;
}
node JZ(node a,node b){ // 矩阵运算，构造矩阵 C
    node c;
    for(int i = 0 ; i <= 9 ; i++)
        for(int j = 0 ; j <= 9 ; j++){
            c.m[i][j] = 0;
            for(int k = 0 ; k <= 9 ; k++)
                c.m[i][j] += (a.m[i][k] * b.m[k][j]) % mod;
            c.m[i][j] %= mod;
    }
    return c;
}
node KSM(node a,node b,int x){ // 矩阵快速幂
    while(x){
        if(x & 1)
            b = JZ(a,b);
        a = JZ(a,a);
        x >>= 1;
    }
    return b;
}
int main()
{
    while(scanf("%lld %lld",&K,&mod) != EOF){
        init();
        for(int i = 0 ; i <= 9 ; i++)
            scanf("%d",&in.m[0][i]);
        if(K < 10) printf("%lld\n",K % mod);
        else{
            node ans = KSM(in,un,K - 9);
            LL cut = 0;
            for(int i = 0 ; i < 10 ; i++)
                cut += (ans.m[0][i] * (9 - i)) % mod;
            printf("%lld\n",cut % mod);
        }
    }
    return 0;
}