【BZOJ3329】Xorequ(数位DP,矩阵乘法)

xiaoxiao2021-02-28  16

Description

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Solution

发现 x3x=2x x ⊕ 3 x = 2 x x2x=3x x ⊕ 2 x = 3 x ,考虑异或是不进位的加法,题目条件等价于 x x 的二进制表示中不存在连续的11。 第一问数位dp,第二问直接矩乘优化dp即可。

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; }
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