HDU 4990 Reading comprehension（矩阵快速幂）题解

思路：

如图找到推导公式，然后一通乱搞就好了

要开long long，否则红橙作伴

代码：

#include<set>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define ll long long
const int maxn = 3;
const int MOD = 1000000000+7;
const int INF = 0x3f3f3f3f;
using namespace std;
ll m;
struct Mat{
    ll s[maxn][maxn];
    void init(){
        for(int i = 0;i < maxn;i++)
            for(int j = 0;j < maxn;j++)
                s[i][j] = 0;
    }
};

Mat mul(Mat a,Mat b){
    Mat t;
    t.init();
    for(int i = 0;i < maxn;i++){
        for(int j = 0;j < maxn;j++){
            for(int k = 0;k < maxn;k++){
                t.s[i][j] = (t.s[i][j] + a.s[i][k]*b.s[k][j])%m;
            }
        }
    }
    return t;
}
Mat pow_mat(Mat p,int n){
    Mat ret;
    ret.init();
    for(int i = 0;i < maxn;i++)
        ret.s[i][i] = 1;
    while(n){
        if(n & 1) ret = mul(ret,p);
        p = mul(p,p);
        n >>= 1;
    }
    return ret;
}
int main(){
    ll n;
    while(scanf("%lld%lld",&n,&m) != EOF){
        Mat A,B,T;
        memset(T.s,0,sizeof(T.s));
        memset(B.s,0,sizeof(B.s));
        T.s[0][0] = T.s[1][0] = T.s[0][2] = T.s[2][2] = 1;
        T.s[0][1] = 2;
        if(n == 1) printf("%lld\n",1%m);
        else if(n == 2) printf("%lld\n",2%m);
        else{
            B.s[0][0] = 2,B.s[1][0] = 1,B.s[2][0] = 1;
            A = pow_mat(T,n - 2);
            A = mul(A,B);
            printf("%lld\n",A.s[0][0]);
        }
    }
    return 0;
}