AtCoder Beginner Contest 397 (OMRON Corporation Programming Contest 2025)
A - Thermometer
题意
思路
模拟
代码
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#include<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 2e5 + 10;
void solve()
{
double x;
cin >> x;
if (x >= 38.0)
{
cout << 1 << endl;
}
else if (x < 37.5)
{
cout << 3 << endl;
}
else
{
cout << 2 << endl;
}
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
//cin >> __;
while (__--)
{
solve();
}
return 0;
}
B - Ticket Gate Log
题意
思路
模拟
代码
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#include<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 2e5 + 10;
void solve()
{
string s;
cin >> s;
int ans = 0;
for (int i = 0; i < s.length(); i++)
{
if (i & 1)
{
if (s[i] != 'o')
{
s.insert(i, "i");
ans++;
}
}
else
{
if (s[i] != 'i')
{
s.insert(i, "o");
ans++;
}
}
}
if (((int)s.size() & 1))
{
ans++;
}
cout << ans << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
//cin >> __;
while (__--)
{
solve();
}
return 0;
}
C - Variety Split Easy
题意
思路
在遍历的同时记录两边不同的数的种类
代码
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#include<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 2e5 + 10;
void solve()
{
int n, ans = -1;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++)
{
cin >> a[i];
}
unordered_map<int, int> l, r;
for (int i = 0; i < n; i++)
{
r[a[i]]++;
}
for (int i = 0; i < n; i++)
{
l[a[i]]++;
r[a[i]]--;
if (r[a[i]] == 0)
{
r.erase(a[i]);
}
ans = max(ans, (int)l.size() + (int)r.size());
}
cout << ans << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
//cin >> __;
while (__--)
{
solve();
}
return 0;
}
D - Cubes
题意
思路
枚举\(a\)，求出\(k\)，并且\(k\)要是正整数，同样，\(\sqrt △\)也要是正整数，由于数值太大，需要借助二分来判断，求出的\(\frac {-a + d(\sqrt △)} 2\)同样要是正整数
代码
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#include<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 2e5 + 10;
int f(int x)
{
if (x < 0)
{
return -1;
}
int l = 0;
int r = 1;
while (r * r < x)
{
r *= 2;
}
while (l <= r)
{
int mid = (l + r) >> 1;
int sq = mid * mid;
if (sq == x)
{
return mid;
}
else if (sq < x)
{
l = mid + 1;
}
else
{
r = mid - 1;
}
}
return -1;
}
void solve()
{
int n;
cin >> n;
for (int a = 1; a * a * a <= n; a++)
{
int temp = n - a * a * a;
if (temp < 0)
{
continue;
}
if (temp % (3 * a) != 0)
{
continue;
}
int k = temp / (3 * a);
int dd = a * a + 4 * k;
int d = f(dd);
if (d == -1)
{
continue;
}
if (d < a || ((d - a) & 1))
{
continue;
}
int y = (d - a) >> 1;
if (y <= 0)
{
continue;
}
int x = a + y;
cout << x << " " << y << endl;
return;
}
cout << -1 << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
//cin >> __;
while (__--)
{
solve();
}
return 0;
}
E - Path Decomposition of a Tree
题意
思路
详情见代码
代码
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#include<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 1e6 + 10;
void solve()
{
int n, k;
cin >> n >> k;
vector<vector<int>> g(n * k);
vector<int> sz(n * k, 0); // sz[i] 代表以 i 为根的子树的大小(结点数)
for (int i = 0; i < n * k - 1; i++)
{
int u, v;
cin >> u >> v;
u--, v--;
g[u].push_back(v);
g[v].push_back(u);
}
if (k == 1)
{
cout << "Yes" << endl;
return;
}
function<void(int, int)> dfs = [&](int u, int fa) {
sz[u] = 1;
int child = 0;
for (auto& v : g[u])
{
if (v == fa)
{
continue;
}
dfs(v, u);
if (sz[v])
{
child++;
sz[u] += sz[v];
}
}
if (sz[u] < k)
{
// 路径还没完成，只能充当非连接点
if (child > 1)
{
cout << "No" << endl;
exit(0);
}
}
else if (sz[u] == k)
{
// 因为要保证“一条”路径，最多两个孩子，此时 u 作连接点
if (child > 2)
{
cout << "No" << endl;
exit(0);
}
sz[u] = 0; // 恰好一条路径，删除无影响
}
else
{
cout << "No" << endl;
exit(0);
}
};
dfs(0, -1);
cout << "Yes" << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
//cin >> __;
while (__--)
{
solve();
}
return 0;
}
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