Japan Registry Services (JPRS) Programming Contest 2024 (AtCoder Beginner Contest 339)
Japan Registry Services (JPRS) Programming Contest 2024 (AtCoder Beginner Contest 339)
A - TLD
代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;
const int N = 1010;
const int mod = 1e9 + 7;
void solve()
{
string s;
cin >> s;
int idx = 0;
int n = s.size();
for (int i = 0; i < n; i++)
{
char c = s[i];
if (c == '.')
{
idx = i;
}
}
cout << s.substr(idx + 1) << endl;
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
B - Langton's 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
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;
const int N = 110;
char g[N][N];
void solve()
{
int n, m, k;
cin >> n >> m >> k;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
g[i][j] = '.';
}
}
int x = 1;
int y = 1;
int f = 0;
while (k--)
{
// cout << x << ' ' << y << endl;
if (g[x][y] == '.')
{
g[x][y] = '#';
f += 1;
f += 4;
f %= 4;
if (f == 0)
{
x = ((x - 1) - 1) % n;
x += n;
x %= n;
x += 1;
}
else if (f == 1)
{
y = y % m;
y += m;
y %= m;
y += 1;
}
else if (f == 2)
{
x = ((x - 1) + 1) % n;
x += n;
x %= n;
x += 1;
}
else
{
y = (y - 2) % m;
y += m;
y %= m;
y += 1;
}
}
else
{
g[x][y] = '.';
f -= 1;
f += 4;
f %= 4;
if (f == 0)
{
x = ((x - 1) - 1) % n;
x += n;
x %= n;
x += 1;
}
else if (f == 1)
{
y = y % m;
y += m;
y %= m;
y += 1;
}
else if (f == 2)
{
x = ((x - 1) + 1) % n;
x += n;
x %= n;
x += 1;
}
else
{
y = (y - 2) % m;
y += m;
y %= m;
y += 1;
}
}
}
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
cout << g[i][j];
}
cout << endl;
}
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
C - Perfect Bus
代码：
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;
const int N = 110;
char g[N][N];
void solve()
{
ll n;
cin >> n;
ll cur = 0;
ll t = 1e18;
for (int i = 1; i <= n; i++)
{
int x;
cin >> x;
cur += x;
t = min(t, cur);
}
if (t >= 0)
{
cout << cur << endl;
}
else
{
cout << cur - t << endl;
}
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
D - Synchronized Players
解题思路:
\(bfs\)搜索，枚举两位棋手从开始位置移动的每一种情况，最多\(60^4\)个位置对。
代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;
const int N = 65;
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
int dist[N][N][N][N];
bool vis[N][N][N][N];
struct node
{
pii a;
pii b;
};
void solve()
{
int n;
cin >> n;
vector<string> g(n + 1);
for (int i = 1; i <= n; i++)
{
cin >> g[i];
g[i] = ' ' + g[i];
}
vector<pii> st;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
if (g[i][j] == 'P')
{
st.push_back({i, j});
}
}
}
memset(dist, 0x3f, sizeof dist);
queue<node> q;
q.push((node){{st[0].fi, st[0].se}, {st[1].fi, st[1].se}});
dist[st[0].fi][st[0].se][st[1].fi][st[1].se] = 0;
while (q.size())
{
auto t = q.front();
q.pop();
int x1 = t.a.fi;
int y1 = t.a.se;
int x2 = t.b.fi;
int y2 = t.b.se;
if (vis[x1][y1][x2][y2])
{
continue;
}
vis[x1][y1][x2][y2] = true;
for (int i = 0; i < 4; i++)
{
int nx1 = max(1, min(x1 + dx[i], n));
int ny1 = max(1, min(y1 + dy[i], n));
int nx2 = max(1, min(x2 + dx[i], n));
int ny2 = max(1, min(y2 + dy[i], n));
if (g[nx1][ny1] == '#')
{
nx1 -= dx[i];
ny1 -= dy[i];
}
if (g[nx2][ny2] == '#')
{
nx2 -= dx[i];
ny2 -= dy[i];
}
dist[nx1][ny1][nx2][ny2] = min(dist[nx1][ny1][nx2][ny2], dist[x1][y1][x2][y2] + 1);
q.push({{nx1, ny1}, {nx2, ny2}});
}
}
ll ans = 1e18;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
ans = min(ans, (ll)dist[i][j][i][j]);
}
}
if (ans >= 1e9)
{
ans = -1;
}
cout << ans << endl;
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
E - Smooth Subsequence
解题思路:
线段树优化\(dp\)。
\(dp[i]:以第i个数为结尾的最长合法序列\)。
对于\(a[i]\)，我们能将结尾数字为\([a[i] - d,a[i] + d]\)的序列和其拼接。
也就是说我们要找到结尾在\([a_i-d,a_i+d]\)这个区间上的最大\(dp\)值，区间查询，单点修改。
代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;
const int N = 5e5 + 10;
int n, d;
int maxv[N * 4];
int a[N];
int dp[N];
void pushup(int u)
{
maxv[u] = max(maxv[u << 1], maxv[u << 1 | 1]);
}
void modify(int u, int l, int r, int x, int val)
{
if (l == r)
{
maxv[u] = max(maxv[u], val);
return;
}
int mid = l + r >> 1;
if (x <= mid)
{
modify(u << 1, l, mid, x, val);
}
else
{
modify(u << 1 | 1, mid + 1, r, x, val);
}
pushup(u);
}
int query(int u, int l, int r, int nl, int nr)
{
if (l >= nl && r <= nr)
{
return maxv[u];
}
int res = 0;
int mid = l + r >> 1;
if (nl <= mid)
{
res = max(res, query(u << 1, l, mid, nl, nr));
}
if (nr > mid)
{
res = max(res, query(u << 1 | 1, mid + 1, r, nl, nr));
}
return res;
}
void solve()
{
cin >> n >> d;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
}
int ans = 0;
int m = N - 1;
for (int i = 1; i <= n; i++)
{
dp[i] = query(1, 1, m, max(1, a[i] - d), min(500000, a[i] + d)) + 1;
ans = max(ans, dp[i]);
modify(1, 1, m, a[i], dp[i]);
}
cout << ans << endl;
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
F - Product Equality
解题思路:
哈希乱搞。
代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;
const int m1 = 1e9 + 7;
const int m2 = 13331;
const int m3 = 1145141;
const int N = 1010;
ll a[N];
ll b[N];
ll d[N];
void solve()
{
int n;
cin >> n;
map<piii, int> q;
for (int i = 1; i <= n; i++)
{
string s;
cin >> s;
for (auto c : s)
{
a[i] = (a[i] * 10 + (c - '0')) % m1;
b[i] = (b[i] * 10 + (c - '0')) % m2;
d[i] = (d[i] * 10 + (c - '0')) % m3;
}
q[{a[i], {b[i], d[i]}}]++;
}
ll ans = 0;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
ll t1 = a[i] * a[j] % m1;
ll t2 = b[i] * b[j] % m2;
ll t3 = d[i] * d[j] % m3;
ans += q[{t1, {t2, t3}}];
}
}
cout << ans << endl;
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
G - Smaller Sum
解题思路:
动态开点 + 主席树。
代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;
const int N = 2e5 + 10;
ll s[N * 65];
int a[N];
int root[N];
int idx = 0;
int ls[N * 65], rs[N * 65];
void insert(auto &pre, auto &cur, int l, int r, int val)
{
cur = ++idx;
s[cur] = s[pre] + val;
if (l == r)
{
return;
}
int mid = l + r >> 1;
if (val <= mid)
{
rs[cur] = rs[pre];
insert(ls[pre], ls[cur], l, mid, val);
}
else
{
ls[cur] = ls[pre];
insert(rs[pre], rs[cur], mid + 1, r, val);
}
}
ll query(int pre, int u, int l, int r, int nl, int nr)
{
if (l >= nl && r <= nr)
{
return s[u] - s[pre];
}
int mid = l + r >> 1;
ll res = 0;
if (nl <= mid)
{
res += query(ls[pre], ls[u], l, mid, nl, nr);
}
if (nr > mid)
{
res += query(rs[pre], rs[u], mid + 1, r, nl, nr);
}
return res;
}
void solve()
{
int n;
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
insert(root[i - 1], root[i], 0, 1e9, a[i]);
}
int q;
cin >> q;
ll ans = 0;
while (q--)
{
ll a, b, c;
cin >> a >> b >> c;
ll l = a ^ ans;
ll r = b ^ ans;
ll x = c ^ ans;
ans = query(root[l - 1], root[r], 0, 1e9, 0, x);
printf("%lld\n", ans);
}
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
浙公网安备 33010602011771号