hdu1757

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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;}

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