A Simple Math Problem（矩阵快速幂）

原创于 2019-08-02 09:41:21 发布 · 608 阅读

0 ·

CC 4.0 BY-SA版权

代码专栏收录该内容

117 篇文章

订阅专栏

该博客讨论了一个简单的数学问题，涉及一个基于特定规则的递归函数f(x)。当x小于10时，f(x)等于x，否则通过一系列系数a0到a9与f(x-1)至f(x-10)的组合来计算。博客提供了输入输出样例，并解释了如何使用矩阵快速幂的方法高效地计算f(k)模m的结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A Simple Math Problem

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

| F0 F1 F2 F3 F4 F5 F6 F7 F8 F9 |
| 1 0 0 0 0 0 0 0 0 0 |
| 0 1 0 0 0 0 0 0 0 0 | 的n次方乘上
| 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 |

| 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 |
| 0 0 0 0 0 0 0 0 0 1 |
即可得出F(n-1)—F(n-9)

即可求出F(n)

ac代码：

#include<iostream>
#include<algorithm>                                
#include<string.h>
using namespace std;
int mod;
struct node{
	int m[15][15];
};
node init,unit;
void vii(){

	memset(init.m ,0,sizeof(init));
	for(int i=1;i<10;i++){
		init.m[i][i-1]=1;
	}
		memset(unit.m ,0,sizeof(unit));
	for(int i=0;i<10;i++){
		unit.m [i][i]=1;
	}
}
node mul(node a,node b){
	node ans;
//	memset(ans.m ,0,sizeof(ans.m ));
	for(int i=0;i<10;i++){
		for(int j=0;j<10;j++){
			ans.m [i][j]=0;
			for(int k=0;k<10;k++){
				ans.m[i][j]=(ans.m [i][j]+a.m [i][k]*b.m [k][j]%mod)%mod;
			}
		}
	}
	return ans;
}

node ksm(node a,node b,int bb){
	while(bb){
		if(bb&1){
			b=mul(a,b);
		}
		bb>>=1;
		a=mul(a,a);
	}
	return b;
}

int main(){
	long long int k;
	while(cin>>k>>mod){
		vii();
		for(int i=0;i<10;i++){
			cin>>init.m [0][i];
		}
		if(k<10){
			cout<<k%mod<<endl;
		}
		else{
			node res=ksm(init,unit,k-9);
			int ans=0;
			for(int i=0;i<10;i++){
				//cout<<res.m [0][i]<<endl;
				ans+=(res.m [0][i]*(9-i))%mod;
			}
			cout<<ans%mod<<endl;
		}
	}
	return 0;
}