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最新推荐文章于 2019-05-14 15:56:22 发布

原创最新推荐文章于 2019-05-14 15:56:22 发布 · 1.3k 阅读

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本文介绍了一种使用矩阵快速幂解决特定类型的递推数列问题的方法。通过构造特定矩阵并利用快速幂运算，可以高效地计算递推数列中任一位置的值。适用于当递推公式涉及到前若干项的线性组合且要求模意义下的结果时。

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题意：当x<10时f(x) = x.当x>=10，f(x)=a0*f(x-1)+a1*f(x-2)+a2*f(x-3)+ …… +a9*f(x-10),求f(k)%m

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
struct node{
    long long m[10][10];
};
long long m;
node P={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,
        0,0,0,0,0,0,0,0,1,0};               //根据表达式构造出矩阵
node I={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,};
node mul(node a,node b){
    int i,j,k;
    node c;
    for(i=0;i<10;i++)
    for(j=0;j<10;j++){
        c.m[i][j]=0;
        for(k=0;k<10;k++)
        c.m[i][j]+=(a.m[i][k]*b.m[k][j])%m;
        c.m[i][j]%=m;
    }
    return c;
}
node quickmod(long long n){
    node a,b;
    a=P;b=I;
    while(n){
        if(n&1)
        b=mul(b,a);
        n/=2;
        a=mul(a,a);
    }
    return b;
}                                           //矩阵快速幂模板
long long a[20];
int main(){
    long long k,i,sum;
    node temp;
    while(cin>>k>>m){
        for(i=0;i<10;i++)
        cin>>a[i];
        if(k<10){
            cout<<k%m<<endl;
            continue;
        }                                   //小于10的时候单独判断
        for(i=0;i<10;i++)
        P.m[0][i]=a[i];
        temp=quickmod(k-9);
        sum=0;
        for(i=0;i<10;i++)
        sum+=(temp.m[0][i]*(9-i))%m;
        sum%=m;
        cout<<sum<<endl;
    }
    return 0;
}