aaaaaaa……aaa(n个)%p的值 (矩阵快速幂)

觉得还是有必要发日志的……否则就在题海中迷失了自己的脚步咯

题目如标题，下面是程序……不要问我矩阵是怎么想的……

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <stack>
#include <vector>
#include <iostream>
#include "algorithm"
using namespace std;
typedef long long LL;
const int MAX=5;
LL a,n,p;
struct Mat{
    LL x,y;
    LL mat[MAX][MAX];
    Mat (){
        x=y=0;
        memset(mat,0,sizeof(mat));
    }
    Mat operator * (const Mat &cc) {
        int i,j,k;
        Mat zt;
        zt.x=x,zt.y=cc.y;
        for (i=1;i<=zt.x;i++)
         for (j=1;j<=cc.y;j++)
          for (k=1;k<=cc.x;k++)
           zt.mat[i][j]=(zt.mat[i][j]+mat[i][k]*cc.mat[k][j])%p;
        return zt;
    }
}m1,m2;
void init(){
    int i,j;
    scanf("%lld%lld%lld",&a,&n,&p);
    m1.x=2,m1.y=1;
    m1.mat[1][1]=m1.mat[2][1]=a;
    m2.x=2,m2.y=2;
    m2.mat[1][1]=10,m2.mat[1][2]=1;
    m2.mat[2][1]=0,m2.mat[2][2]=1;
}
Mat ksm(Mat zt,LL k){
    Mat an;
    LL i,j;
    an.x=zt.x,an.y=zt.y;
    for (i=1;i<=an.x;i++)
     for (j=1;j<=an.y;j++)
      an.mat[i][j]=(i==j);
    while (k)
    {if (k%2==1)
      an=an*zt;
     zt=zt*zt;
     k/=2;
    }
    return an;
}
int main(){
    freopen ("ksm.in","r",stdin);
    freopen ("ksm.out","w",stdout);
    int i,j;
    init();
    m1=ksm(m2,n-1)*m1;
    printf("%lld",m1.mat[1][1]);
    return 0;
}