前四题都好水。后面两道题好难。
#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; }
题意：给一个有向图，问能不能从$S$，走$3^{x}$ $x \geq 1$ 能输出$x$，不能输出-1
思路：BFS。刚开始想的是对每一个点，枚举它往后走三步的点，但是T了。正解应该用$dis\left[ i\right] \left[ k\right]$表示走到$i$，并且走的步数模3等于$k$ 然后BFS就OK了。
#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; }
题意：给定一个$N$,$K$，问长度为$K$，且相邻两数的乘积不超过$N$的方案数。
思路：我只会$O\left( KN^{2}\right)$的暴力。看了题解瞬间觉得...tql
将数分为小于等于$\sqrt {N}$和大于$\sqrt {N}$，记为$s$和$b$
把$b$分成$\sqrt {N}$块 第$i$块表示$i\cdot b\leq N$ 那么这个块里有$\dfrac {N}{i}-\dfrac {N}{i+1}$个数
$S\left( i,j\right)$表示长度为$i$，最后一个数为$j$ ($j\leq \sqrt {N}$)的方案数
$B\left( i,j\right)$表示长度为$i$，最后一个数在块$B^{j}$里($j\leq \sqrt {N}$)的方案数
一个小的数前面可以放任意一个小的数，也可以放一个大数，这个数与这个小的数乘积小于等于$N$
那么$S$的递推式子是$S\left( i,j\right) =\sum ^{\sqrt {N}}_{k=1}s\left( i-1,k\right) +\sum ^{\sqrt {N}}_{k=j}B\left( i-1,k\right)$
$B$的递推式子是$B\left( i,j\right) =\left( \dfrac {N}{j}-\dfrac {N}{j+1}\right) \sum ^{j}_{k=1}S\left( i-1,k\right)$
$ans=\sum ^{\sqrt {N}}_{i=1}\left( S\left( k,i\right) +B\left( k,i\right) \right)$
但是这样复杂度是$O\left( k\left( \sqrt {N}\right) ^{2}\right)$
在计算的过程中计算完这一层可以把这一层的给求个和，复杂度就降为$O\left( k\sqrt {N}\right)$
好题。
#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; }