A Simple Math Problem

最新推荐文章于 2020-11-19 20:12:12 发布

踢到火星

最新推荐文章于 2020-11-19 20:12:12 发布

阅读量449

点赞数

CC 4.0 BY-SA版权

分类专栏：数论

本文链接：https://blog.youkuaiyun.com/qq_40941611/article/details/81269796

数论专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一种使用矩阵快速幂解决特定递推数列问题的方法。对于给定的递推公式f(x)，当x小于10时，f(x) = x；当x大于等于10时，f(x)为前10项的加权和。文章提供了完整的C++代码实现，并解释了如何构造关系矩阵以及如何应用快速幂算法高效地计算f(k) % m。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

思路：这个是矩阵快速幂，没看出来的话不好做要是看出来就很好做了。

关系矩阵是 a0 a1 a2 a3 a4 a5 a6 a7 a8 a9

1 0 0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 1 0 0

0 0 0 0 0 0 0 0 1 0

(矩阵快速幂一直不会写，只会直接粘贴代码 - -）

#include <iostream>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <cmath>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
int n;
long long k,m;
int N;
struct node{long long  a[11][11];};
node shu,ans,mp;
node matrix(node x,node y){
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            mp.a[i][j]=0;
            for(int p=1;p<=n;p++)
                mp.a[i][j]=(mp.a[i][j]+x.a[i][p] * y.a[p][j])%N;
        }
    return mp;
}
void work(long long k){
    while(k){
        if(k&1)
            ans=matrix(ans,shu);
        k>>=1;
        shu=matrix(shu,shu);
    }
}
int main(){
    while(~scanf("%lld%d",&k,&N))
    {
    n=10;
//    sum=0;
    memset(ans.a,0,sizeof(ans.a));
    memset(shu.a,0,sizeof(shu.a));
    for(int i=1;i<=n;i++){
        ans.a[i][i]=1;
    }
    for(int i=1;i<=n;i++)
    {
        cin>>shu.a[1][i];
    }
    int t=1;
    for(int i=2;i<=n;i++)
    {
        shu.a[i][t]=1;
        t++;
    }
    work(k-9);
    long long sum=0;
    t=9;
    for(int i=1;i<=n;i++){
        sum+=ans.a[1][i]*t;
        sum=sum%N;
        t--;
    }
    cout<<sum<<endl;
    }
    return 0;
}