HDU 5950 Recursive sequence（矩阵快速幂）题解

思路：一开始不会n^4的推导，原来是要找n和n-1的关系，这道题的MOD是long long 的，矩阵具体如下所示

最近自己总是很坑啊，代码都瞎吉坝写，一个long long的输入写成%d一直判我TLE，一度怀疑矩阵快速幂地复杂度orz

代码：

#include<set>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define ll long long
const int maxn = 7;
const ll MOD = 2147493647;
const int INF = 0x3f3f3f3f;
using namespace std;
struct Mat{
    ll s[maxn][maxn];
};

Mat mul(Mat a,Mat b){
    Mat t;
    memset(t.s,0,sizeof(t));
    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])%MOD;
            }
        }
    }
    return t;
}
Mat pow_mat(Mat p,int n){
    Mat ret;
    memset(ret.s,0,sizeof(ret.s));
    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 T,a,b,n;
    Mat A,B,C;
    memset(A.s,0,sizeof(A.s));
    A.s[0][0] = A.s[0][2] = A.s[1][0] = A.s[2][2] = A.s[3][3] = A.s[4][4] = A.s[5][5] = 1;
    for(int i = 0;i < 7;i++)
        A.s[i][6] = 1;
    A.s[1][6] = 0;
    A.s[0][1] = A.s[4][5] = 2;
    A.s[3][4] = A.s[3][5] = 3;
    A.s[0][3] = A.s[0][5] = A.s[2][3] = A.s[2][5] = 4;
    A.s[0][4] = A.s[2][4] = 6;
    /*for(int i = 0;i < 7;i++){
        for(int j = 0;j < 7;j++){
            printf("%d ",A.s[i][j]);
        }
        printf("\n");
    }*/
    scanf("%lld",&T);
    while(T--){
        scanf("%lld%lld%lld",&n,&a,&b);
        if(n == 1){
            printf("%lld\n",a);
        }
        else if(n == 2){
            printf("%lld\n",b);
        }
        else{
            C = pow_mat(A,n - 2);
            ll ans = 0;
            ans = (C.s[0][0]*b%MOD + C.s[0][1]*a%MOD + C.s[0][2]*16%MOD +
                   C.s[0][3]*8%MOD + C.s[0][4]*4%MOD + C.s[0][5]*2%MOD + C.s[0][6])%MOD;
            printf("%lld\n",ans);
        }
    }
    return 0;
}