The first four questions are good water. Behind two very difficult questions.
#include <cstdio> #include <algorithm> using namespace std; inline int read() { int x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); } return x * f; } const int N = 1e5 + 10; int a[N]; int main() { int n = read(); for (int i = 0; i < n; i++) a[i] = read(); sort(a, a + n); printf("%d\n", a[n / 2] - a[n / 2 - 1]); return 0; }
#include <cstdio> using namespace std; inline int read() { int x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); } return x * f; } const int MOD = 1e9 + 7; const int N = 2010; int C[N][N]; void init() { C[0][0] = 1; C[1][0] = C[1][1] = 1; for (int i = 2; i < N; i++) { C[i][0] = 1; for (int j = 1; j <= i; j++) C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD; } } int main() { init(); int n = read(), k = read(); for (int i = 1; i <= k; i++) { printf("%d\n", 1LL * C[k - 1][i - 1] * C[n - k + 1][i] % MOD); } return 0; }
Question is intended: to a directed graph can ask from $ S $, take $ 3 ^ {x} $ $ x \ geq 1 $ $ X $ can be output, the output is not -1
Ideas: BFS. At first thought was for each point, enumerate it take three steps backward point, but the T. Positive solutions should $ dis \ left [i \ right] \ left [k \ right] $ denotes come $ i $, and take the number of steps equal to the mold 3 and $ k $ BFS OK.
#include <bits/stdc++.h> using namespace std; inline int read() { int x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); } return x * f; } const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; int n, m, s, t, dis[N][3]; vector<int> G[N]; void bfs() { memset(dis, -1, sizeof(dis)); dis[s][0] = 0; queue< pair<int, int> > que; que.push({s, 0}); while (!que.empty()) { pair<int, int> u = que.front(); que.pop(); for (auto v : G[u.first]) { if (dis[v][(u.second + 1) % 3] < 0) { dis[v][(u.second + 1) % 3] = dis[u.first][u.second] + 1; que.push({v, (u.second + 1) % 3}); } } } } int main() { n = read(), m = read(); while (m--) { int u = read(), v = read(); G[u].push_back(v); } s = read(), t = read(); bfs(); if (dis[t][0] == -1) puts("-1"); else printf("%d\n", dis[t][0] / 3); return 0; }
Meaning of the questions: Given a $ N $, $ K $, ask a length of $ K $, and the number of programs the product of two numbers do not exceed $ N $ neighbor.
Thinking: I can only $ O \ violence left (KN ^ {2} \ right) $ a. Read an instant solution to a problem that ... tql
The less the number of divided $ \ sqrt {N} $ greater than $ \ sqrt {N} $, denoted by $ s $ and $ b $
The $ b $ into $ \ sqrt {N} $ block of $ I $ block represents $ i \ cdot b \ leq N $ then this block there $ \ dfrac {N} {i} - \ dfrac {N} {i $ the number of + 1}
$ S \ left (i, j \ right) $ denotes the length of $ i $, last number $ j $ ($ j \ leq \ sqrt {N} $) the number of programs
$ B \ left (i, j \ right) $ denotes the length of $ i $, $ B number in the last block of $ ^ {j} where ($ j \ leq \ sqrt {N} $) scheme
A small number of any of the foregoing may be put in a small number, a large number may be put, the number of small numbers of this product is less than equal to $ N $
Then $ S $ recursive expression is $ S \ left (i, j \ right) = \ sum ^ {\ sqrt {N}} _ {k = 1} s \ left (i-1, k \ right) + \ sum ^ {\ sqrt {N}} _ {k = j} B \ left (i-1, k \ right) $
$ B $ recursive formula is $ B \ left (i, j \ right) = \ left (\ dfrac {N} {j} - \ dfrac {N} {j + 1} \ right) \ sum ^ { j} _ {k = 1} S \ left (i-1, k \ right) $
$ans=\sum ^{\sqrt {N}}_{i=1}\left( S\left( k,i\right) +B\left( k,i\right) \right)$
But this complexity is $ O \ left (k \ left (\ sqrt {N} \ right) ^ {2} \ right) $
After the calculation in the calculation of this layer of this layer can give a request and, on the reduced complexity $ O \ left (k \ sqrt {N} \ right) $
Good question.
#include <cstdio> #include <cstring> #include <cmath> #define ll long long using namespace std; const ll MOD = 1e9 + 7; const int N = 5e4 + 10; ll S[110][N], B[110][N]; inline int read() { int x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); } return x * f; } int main() { int n = read(), k = read(); int r = sqrt(n) + 1; S[0][1] = 1; for (int i = 0; i < k; i++) { for (int j = 1; j < r; j++) S[i][j] = (S[i][j] + S[i][j - 1]) % MOD; for (int j = r - 2; j > 0; j--) B[i][j] = (B[i][j] + B[i][j + 1]) % MOD; for (int j = 1; j < r; j++) { B[i + 1][j] = S[i][j] * (n / j - n / (j + 1)); if (j == n / j) B[i + 1][j] = 0; //已被包含在S里 B[i + 1][j] %= MOD; } for (int j = 1; j < r; j++) { S[i + 1][j] = B[i][j] + S[i][r - 1]; S[i + 1][j] %= MOD; } } ll ans = 0; for (int i = 1; i <= r; i++) ans = (ans + B[k][i] + S[k][i]) % MOD; printf("%lld\n", ans); return 0; }