AtCoder Beginner Contest 335 (Sponsored by Mynavi)
AtCoder Beginner Contest 335 (Sponsored by Mynavi)
A - 2023
代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
void solve()
{
string s;
cin >> s;
int n = s.size() - 1;
s[n] = '4';
cout << s << endl;
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
B - Tetrahedral Number
代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
void solve()
{
int n;
cin >> n;
for (int i = 0; i <= n; i++)
{
for (int j = 0; j <= n; j++)
{
for (int k = 0; k <= n; k++)
{
if (i + j + k <= n)
{
cout << i << ' ' << j << ' ' << k << endl;
}
}
}
}
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
C - Loong Tracking
代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
void solve()
{
int n, q;
cin >> n >> q;
vector<pii> v;
for (int i = 1; i <= n; i++)
{
v.push_back({i, 0});
}
reverse(v.begin(), v.end());
int dx = 0;
int dy = 0;
while (q--)
{
int t;
cin >> t;
if (t == 1)
{
string s;
cin >> s;
auto t = v.back();
if (s[0] == 'U')
{
v.push_back({t.fi, t.se + 1});
}
else if (s[0] == 'D')
{
v.push_back({t.fi, t.se - 1});
}
else if (s[0] == 'L')
{
v.push_back({t.fi - 1, t.se});
}
else
{
v.push_back({t.fi + 1, t.se});
}
}
else
{
int x;
cin >> x;
int idx = v.size() - 1 - x + 1;
cout << v[idx].fi << ' ' << v[idx].se << endl;
}
}
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
D - Loong and Takahashi
代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
const int N = 50;
int d[N][N];
bool st[N][N];
int n;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int cnt = 0;
bool check(int a, int b)
{
if (a < 1 || a > n || b < 1 || b > n)
{
return false;
}
if (d[a][b])
{
return false;
}
return true;
}
void dfs(int a, int b, int t)
{
// cout << a << ' ' << b << endl;
auto cur = (t + 1) % 4;
int x = a + dx[cur];
int y = b + dy[cur];
if (check(x, y))
{
d[x][y] = ++cnt;
dfs(x, y, t);
}
else
{
if (d[x][y] == -1)
{
return;
}
dfs(a, b, cur);
}
}
void solve()
{
cin >> n;
int idx = (n + 1) / 2;
// st[idx][idx] = true;
d[idx][idx] = -1;
d[1][1] = ++cnt;
dfs(1, 1, -1);
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
if (d[i][j] == -1)
{
cout << "T" << ' ';
}
else
{
cout << d[i][j] << ' ';
}
}
cout << endl;
}
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
E - Non-Decreasing Colorful Path
解题思路：
\(bfs\)。每次先更新权值最小的点。因为大权值点不可能回搜小权值，所以每个点的距离最多被遍历一次。
代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
void solve()
{
int n, m;
cin >> n >> m;
vector<vector<int>> adj(n + 1, vector<int>());
vector<int> a(n + 1);
for (int i = 1; i <= n; i++)
{
cin >> a[i];
}
for (int i = 1; i <= m; i++)
{
int u, v;
cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
priority_queue<pii> q;
vector<int> d(n + 1);
q.push({-a[1], 1});
d[1] = 1;
while (q.size())
{
auto t = q.top();
int u = t.se;
// cout << t.fi << endl;
q.pop();
for (auto v : adj[u])
{
if (a[v] >= a[u])
{
if (d[v] < d[u] + (a[v] != a[u]))
{
d[v] = d[u] + (a[v] != a[u]);
q.push({-a[v], v});
}
}
}
}
cout << d[n] << endl;
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
F - Hop Sugoroku
解题思路：
\(f[i]:第i点为黑点的方案数\)。
暴力更新代码:
for(int i = 1;i<=n;i++)
{
for(int j = i + a[i];j <= n;j += a[i])
{
f[j] = (f[j] + f[i]) % mod;
}
}
很明显，时间复杂度为\(O(n^2)\)。爆了。
我们发现，对于\(a[i] > \sqrt{N}\)的遍历，时间复杂度为\(O(n\sqrt{n})\)，不会爆。
所以，根号分治。
我们想办法处理\(a[i] < \sqrt{N}\)的情况，这样的\(a[i]\)数量不超过\(\sqrt{N}\)个。
对于位置\(i\)，我们能够更新到的位置为\(j = i + k \times a[i],\ j > i\)。
可以发现\(i \%a[i] == j \% a[i]\)。
\(g[a[i]][i \%a[i]:表示当前(idx\%a[i] == i\% a[i])的累加量\)。
即对于\(f[1] \sim f[cur]\),若\(a[i] \leq \sqrt{N}\)，\(g[a[i]][i \% a[i]] = (g[a[i]][i\% a[i]] + f[i]) \% mod\)。
对于当前\(f[i]\),将所有能从前面转移过来的情况加上。
for(int j = 1;j <= sqrt(N);j ++)
{
f[i] = (f[i] + g[j][i % j]) % mod;
}
代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
const int N = 2e5;
const int mod = 998244353;
ll g[500][500];
void solve()
{
int n;
cin >> n;
vector<ll> a(n + 1);
for (int i = 1; i <= n; i++)
{
cin >> a[i];
}
int lim = sqrt(N) + 1;
vector<ll> f(n + 1);
f[1] = 1;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= lim; j++)
{
f[i] = (f[i] + g[j][i % j]) % mod;
}
if (a[i] > lim)
{
for (int j = i + a[i]; j <= n; j += a[i])
{
f[j] = (f[j] + f[i]) % mod;
}
}
else
{
g[a[i]][i % a[i]] = (g[a[i]][i % a[i]] + f[i]) % mod;
}
}
ll ans = 0;
for (int i = 1; i <= n; i++)
{
ans = (ans + f[i]) % mod;
}
cout << ans << endl;
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
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