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a[1]=a[2]=a[3]=1
a[x]=a[x-3]+a[x-1] (x>3)
求a数列的第n项对1000000007（10^9+7）取余的值。
第一行一个整数T，表示询问个数。
以下T行，每行一个正整数n。
每行输出一个非负整数表示答案。
36810
4919
对于30%的数据 n<=100；
对于60%的数据 n<=2*10^7；
对于100%的数据 T<=100，n<=2*10^9；
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include //#include //#pragma GCC optimize(2)using namespace std;#define maxn 200005#define inf 0x7fffffff//#define INF 1e18#define rdint(x) scanf("%d",&x)#define rdllt(x) scanf("%lld",&x)#define rdult(x) scanf("%lu",&x)#define rdlf(x) scanf("%lf",&x)#define rdstr(x) scanf("%s",x)#define mclr(x,a) memset((x),a,sizeof(x))typedef long long ll;typedef unsigned long long ull;typedef unsigned int U;#define ms(x) memset((x),0,sizeof(x))const long long int mod = 1e9 + 7;#define Mod 1000000000#define sq(x) (x)*(x)#define eps 1e-5typedef pair pii;#define pi acos(-1.0)//const int N = 1005;#define REP(i,n) for(int i=0;i<(n);i++)typedef pair pii;inline int rd() { int x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x;}ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b);}int sqr(int x) { return x * x; }/*ll ans;ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans;}*/struct Mat { int a[5][5]; void init() { ms(a); for (int i = 0; i < 5; i++)a[i][i] = 1; }};Mat mul(Mat a, Mat b) { Mat ans; for (int i = 0; i < 5; i++) { for (int j = 0; j < 5; j++) { ans.a[i][j] = 0; for (int k = 0; k < 5; k++) { ans.a[i][j] = (ans.a[i][j] + a.a[i][k] % mod*b.a[k][j] % mod) % mod; ans.a[i][j] %= mod; } } } return ans;}Mat qpow(Mat a, int n) { Mat ans; ans.init(); while (n) { if (n & 1)ans = mul(ans, a); a = mul(a, a); n >>= 1; } return ans;}int main(){ // ios::sync_with_stdio(0); int T = rd(); while (T--) { int n = rd(); if (n == 1 || n == 2 || n == 3)cout << 1 << endl; else { Mat tmp; ms(tmp.a); tmp.a[0][0] = 1; tmp.a[0][2] = 1; tmp.a[1][0] = 1; tmp.a[2][1] = 1; tmp = qpow(tmp, n - 3); ll ans = (1ll * tmp.a[0][0] % mod + 1ll * tmp.a[0][1] % mod + 1ll * tmp.a[0][2] % mod) % mod; printf("%lld\n", ans%mod); } } return 0;}
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