题目大意
略。
分析
关于为什么是卡特兰数，可以这么想，对于经典问题“N个数入栈，有多少种出栈顺序？”，这个问题的答案是卡特兰数的第 N 项，那么本题就可以这么想，“N个数出栈，有多少种入栈顺序？”，这是因为中序遍历就是递归，就是要入栈出栈的。
代码如下
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 5 #define Rep(i,n) for (int i = 0; i < (int)(n); ++i) 6 #define For(i,s,t) for (int i = (int)(s); i <= (int)(t); ++i) 7 #define rFor(i,t,s) for (int i = (int)(t); i >= (int)(s); --i) 8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 12 13 #define pr(x) cout << #x << " = " << x << " " 14 #define prln(x) cout << #x << " = " << x << endl 15 16 #define LOWBIT(x) ((x)&(-x)) 17 18 #define ALL(x) x.begin(),x.end() 19 #define INS(x) inserter(x,x.begin()) 20 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end()) 21 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 22 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower); 23 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper); 24 25 #define ms0(a) memset(a,0,sizeof(a)) 26 #define msI(a) memset(a,0x3f,sizeof(a)) 27 #define msM(a) memset(a,-1,sizeof(a)) 28 29 #define MP make_pair 30 #define PB push_back 31 #define ft first 32 #define sd second 33 34 template<typename T1, typename T2> 35 istream &operator>>(istream &in, pair<T1, T2> &p) { 36 in >> p.first >> p.second; 37 return in; 38 } 39 40 template<typename T> 41 istream &operator>>(istream &in, vector<T> &v) { 42 for (auto &x: v) 43 in >> x; 44 return in; 45 } 46 47 template<typename T> 48 ostream &operator<<(ostream &out, vector<T> &v) { 49 Rep(i, v.size()) out << v[i] << " "[i == v.size() - 1]; 50 return out; 51 } 52 53 template<typename T1, typename T2> 54 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 55 out << "[" << p.first << ", " << p.second << "]" << " "; 56 return out; 57 } 58 59 inline int gc(){ 60 static const int BUF = 1e7; 61 static char buf[BUF], *bg = buf + BUF, *ed = bg; 62 63 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 64 return *bg++; 65 } 66 67 inline int ri(){ 68 int x = 0, f = 1, c = gc(); 69 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 70 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 71 return x*f; 72 } 73 74 template<class T> 75 inline string toString(T x) { 76 ostringstream sout; 77 sout << x; 78 return sout.str(); 79 } 80 81 inline int toInt(string s) { 82 int v; 83 istringstream sin(s); 84 sin >> v; 85 return v; 86 } 87 88 //min <= aim <= max 89 template<typename T> 90 inline bool BETWEEN(const T aim, const T min, const T max) { 91 return min <= aim && aim <= max; 92 } 93 94 typedef unsigned int uI; 95 typedef long long LL; 96 typedef unsigned long long uLL; 97 typedef vector< int > VI; 98 typedef vector< bool > VB; 99 typedef vector< char > VC; 100 typedef vector< double > VD; 101 typedef vector< string > VS; 102 typedef vector< LL > VL; 103 typedef vector< VI > VVI; 104 typedef vector< VB > VVB; 105 typedef vector< VS > VVS; 106 typedef vector< VL > VVL; 107 typedef vector< VVI > VVVI; 108 typedef vector< VVL > VVVL; 109 typedef pair< int, int > PII; 110 typedef pair< LL, LL > PLL; 111 typedef pair< int, string > PIS; 112 typedef pair< string, int > PSI; 113 typedef pair< string, string > PSS; 114 typedef pair< double, double > PDD; 115 typedef vector< PII > VPII; 116 typedef vector< PLL > VPLL; 117 typedef vector< VPII > VVPII; 118 typedef vector< VPLL > VVPLL; 119 typedef vector< VS > VVS; 120 typedef map< int, int > MII; 121 typedef unordered_map< int, int > uMII; 122 typedef map< LL, LL > MLL; 123 typedef map< string, int > MSI; 124 typedef map< int, string > MIS; 125 typedef set< int > SI; 126 typedef stack< int > SKI; 127 typedef deque< int > DQI; 128 typedef queue< int > QI; 129 typedef priority_queue< int > PQIMax; 130 typedef priority_queue< int, VI, greater< int > > PQIMin; 131 const double EPS = 1e-8; 132 const LL inf = 0x7fffffff; 133 const LL infLL = 0x7fffffffffffffffLL; 134 const LL mod = 1e9 + 7; 135 const int maxN = 1e6 + 7; 136 const LL ONE = 1; 137 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 138 const LL oddBits = 0x5555555555555555; 139 140 LL N; 141 142 LL fac[maxN << 1]; 143 void init_fact() { 144 fac[0] = 1; 145 For(i, 1, (N << 1)) { 146 fac[i] = (i * fac[i - 1]) % mod; 147 } 148 } 149 150 //ax + by = gcd(a, b) = d 151 // 扩展欧几里德算法 152 /** 153 * a*x + b*y = 1 154 * 如果ab互质，有解 155 * x就是a关于b的逆元 156 * y就是b关于a的逆元 157 * 158 * 证明： 159 * a*x % b + b*y % b = 1 % b 160 * a*x % b = 1 % b 161 * a*x = 1 (mod b) 162 */ 163 inline void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d){ 164 if (!b) {d = a, x = 1, y = 0;} 165 else{ 166 ex_gcd(b, a % b, y, x, d); 167 y -= x * (a / b); 168 } 169 } 170 171 // 求a关于p的逆元，如果不存在，返回-1 172 // a与p互质，逆元才存在 173 inline LL inv_mod(LL a, LL p = mod){ 174 LL d, x, y; 175 ex_gcd(a, p, x, y, d); 176 return d == 1 ? (x % p + p) % p : -1; 177 } 178 179 inline LL comb_mod(LL m, LL n) { 180 LL ret; 181 182 if(m > n) swap(m, n); 183 184 ret = (fac[n] * inv_mod(fac[m], mod)) % mod; 185 ret = (ret * inv_mod(fac[n - m], mod)) % mod; 186 187 return ret; 188 } 189 190 // 求卡特兰数 191 inline LL Catalan(LL x) { 192 if(x == 0) return 1; 193 LL ret = inv_mod(x + 1, mod) * comb_mod(x, 2 * x); 194 return ret % mod; 195 } 196 197 int main(){ 198 //freopen("MyOutput.txt","w",stdout); 199 //freopen("input.txt","r",stdin); 200 //INIT(); 201 scanf("%lld", &N); 202 init_fact(); 203 if(N < 1) N = 0; 204 printf("%lld ", Catalan(N)); 205 return 0; 206 }