【BZOJ3329】Xorequ（数位DP，矩阵乘法）

Description

Solution

Source

/************************************************
* Au: Hany01
* Date: Feb 25th, 2018
* Prob: BZOJ3329 Xorequ
* Email: hany01@foxmail.com
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define fir first
#define sec second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
register int _, __; register char c_;
for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
return _ * __;
}

inline void File()
{
#ifdef hany01
freopen("bzoj3329.in", "r", stdin);
freopen("bzoj3329.out", "w", stdout);
#endif
}

LL dp[110][2];
int cnt, a[110];

inline void Init()
{
dp[0][0] = 1;
For(i, 1, 100)
dp[i][0] = dp[i - 1][0] + dp[i - 1][1], dp[i][1] = dp[i - 1][0];
}

inline LL dfs(int pos, int pre, int lmt)
{
if (!pos) return 1;
if (!lmt) return dp[pos + 1][pre];
register LL tmp = 0;
For(i, 0, lmt ? a[pos] : 1) if (!pre || !i) tmp += dfs(pos - 1, i, lmt && i == a[pos]);
return tmp;
}

inline LL Solve1(LL n)
{
cnt = 0;
while (n) a[++ cnt] = n % 2, n /= 2;
return dfs(cnt, 0, 1);
}

struct Matrix
{
int ret[2][2], x, y;
Matrix(int a) {
Set(ret, 0);
if (a == 1) ret[0][0] = ret[1][1] = 1;
x = y = 2;
}
};
Matrix operator * (Matrix A, Matrix B) {
Matrix C(0);
rep(i, A.x) rep(j, B.y) rep(k, A.y)
(C.ret[i][j] += A.ret[i][k] * 1ll * B.ret[k][j] % Mod) %= Mod;
return C;
}
Matrix operator ^ (Matrix a, LL b) {
Matrix Ans(1);
for ( ; b; b >>= 1, a = a * a) if (b & 1) Ans = Ans * a;
return Ans;
}

inline int Solve2(LL n)
{
Matrix A(0), B(0);
A.ret[0][0] = 1, A.ret[0][1] = 0, A.ret[1][0] = 1, A.ret[1][1] = 1,
B.ret[0][0] = 1, B.ret[0][1] = 1, B.ret[1][0] = 1, B.ret[1][1] = 0;
B = B ^ (n - 1), A = A * B;
return (A.ret[1][1] + A.ret[1][0]) % Mod;
}

int main()
{
File();
Init();
register LL n;
for (register int T = read(); T --; ) {
scanf("%lld", &n);
printf("%lld\n%d\n", Solve1(n) - 1, Solve2(n));
}
return 0;
}