nuc Magic bracelet of the 2nd Shanxi College Student Programming Contest

problem

It's wonderful... It is found that each row of the state transition matrix can be obtained by translating the previous row. Each time only the first row is counted and then translated, \(O(n^3)\) becomes \(O(n^2 )\) .

#include <iostream>
#include <cstdio>
using namespace std;
int n, k;
struct Matrix{
    int num[205][205];
    Matrix operator*(const Matrix &x)const{
        Matrix re;
        for(int i=1; i<=n; i++){
            re.num[1][i] = 0;
            for(int j=1; j<=n; j++)
                re.num[1][i] = (re.num[1][i]+num[1][j]*x.num[j][i])%100;
        }
        for(int i=2; i<=n; i++)
            for(int j=1; j<=n; j++)
                re.num[i][j] = re.num[i-1][(j-2+n)%n+1];
        return re;
    }
}dan, yua, zhu;
int main(){
    cin>>n>>k;
    for(int i=1; i<=n; i++){
        scanf("%d", &yua.num[1][i]);
        dan.num[i][i] = 1;
        zhu.num[i][i] = zhu.num[i][(i-2+n)%n+1] = 1;
    }
    while(k){
        if(k&1) dan = dan * zhu;
        zhu = zhu * zhu;
        k >>= 1;
    }
    yua = yua * dan;
    for(int i=1; i<=n; i++)
        printf("%d ", yua.num[1][i]);
    printf("\n");
    return 0;
}