Codeforces Round #118 (Div. 1) A. Plant

题目链接：​​Plant​​

题目大意：给你一些图形，问第n个图形里面有几个上三角形

题目思路：下一张图的上三角形和下三角形个数可以由上一张图得到，具体为upn=3∗upn−1+downn−1和downn=3∗downn−1+upn−1，然后构造矩阵就好了

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;
typedef long long ll;
const ll maxn = 2;
const ll MOD = 1e9+7;
#define mod(x) ((x)%MOD)

struct mat{
    ll m[maxn][maxn];
}unit;

mat operator *(mat a,mat b){
    mat ret;
    ll x;
    for(ll i = 0;i < maxn;i++){
        for(ll j = 0;j < maxn;j++){
            x = 0;
            for(ll k = 0;k < maxn;k++)
                x += mod((ll)a.m[i][k]*b.m[k][j]);
            ret.m[i][j] = mod(x);
        }
    }
    return ret;
}

void init_unit(){
    for(ll i = 0;i < maxn;i++)
        unit.m[i][i] = 1;
    return ;
}

mat pow_mat(mat a,ll n){
    mat ret = unit;
    while(n){
        if(n&1) ret = ret*a;
        a = a*a;
        n >>= 1;
    }
    return ret;
}

ll quick_pow(ll a,ll n){
    ll ans = 1;
    while(n){
        if(n&1) ans = (ans*a)%MOD;
        a = (a*a)%MOD;
        n >>= 1;
    }
    return ans;
}

int main(){
    ll n;
    init_unit();
    while(~scanf("%lld",&n)){
        if(n == 0ll) puts("1");
        else if(n == 1ll) puts("3");
        else{
            mat a,b;
            b.m[0][0] = 3,b.m[0][1] = 1;
            b.m[1][0] = 1,b.m[1][1] = 3;


            a.m[0][0] = 3,a.m[0][1] = 1;
            b = pow_mat(b,n-1);
            a = a*b;
            printf("%lld\n",(a.m[0][0]+MOD)%MOD);
        }
    }
    return 0;
}