常用模板
快读：
1 inline char gc() { 2 static char buf[100000], *p1 = buf, *p2 = buf; 3 return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++; 4 } 5 6 //#define gc getchar 7 inline int read() { 8 int x = 0; char ch = gc(); bool f = 1; 9 for (; !isdigit(ch); ch = gc()) if (ch == '-') f = -1; 10 for (; isdigit(ch); ch = gc()) x = x * 10 + ch - '0'; 11 return f * x; 12 }
康托展开:
1 LL fac[20]; 2 3 void init() { 4 fac[0] = 1; 5 for (int i = 1; i <= 15; ++i) fac[i] = fac[i - 1] * i; 6 } 7 8 LL cantor(int a[]) { 9 LL res = 0; 10 for (int i = 1; i <= 8; ++i) { 11 int cnt = 0; 12 for (int j = i + 1; j <= 8; ++j) 13 cnt += a[j] < a[i]; 14 res += cnt * fac[8 - i]; 15 } 16 return res; 17 } 18 19 void inv_cantor(int n, LL k) { 20 vector<int> vec; k--; 21 int a[20]; 22 for (int i = 1; i <= n; ++i) vec.push_back(i); 23 for (int i = n; i >= 1; --i) { 24 LL p = k / fac[i - 1]; k %= fac[i - 1]; 25 sort(vec.begin(), vec.end()); 26 a[n - i + 1] = vec[p]; 27 vec.erase(vec.begin() + t); 28 } 29 for (int i = 1; i <= n; ++i) printf("%d ", a[i]); 30 }
幂次求和：
Σ (i, 1, n) (i) = n * (n + 1) / 2 = 1 / 2 * n2 + 1 / 2 * n
Σ (i, 1, n) (i2) = n * (n + 1) * (2n + 1) / 6 = 1 / 3 * n3 + 1 / 2 * n2 + 1 / 6 * n
Σ (i, 1, n) (i3) = (n * (n + 1) / 2)2 = 1 / 4 * n4 + 1 / 2 * n3 + 1 / 4 * n2
Σ (i, 1, n) (i4) = n * (n + 1) * (2n + 1) * (3x2 + 3x - 1) / 30 = 1 / 5 * n5 + 1 / 2 * n4 + 1 / 3 * n3 - 1 / 30 * n
Σ (i, 1, n) (i4) = n2 * (n + 1) * (2n3 + 4n2 + n - 1) / 12
海伦公式:
S = sqrt(p * (p - a) * (p - b) * (p - c));
p = (a + b + c) / 2.0;
