AtCoder Beginner Contest 400

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A - ABC400 Party

题意

思路

模拟

代码

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#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;

const int mxn = 2e5 + 10;

void solve()
{
	int n;
	cin >> n;
	if (400 % n)
	{
		cout << -1 << endl;
		return;
	}
	cout << 400 / n << 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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B - Sum of Geometric Series

题意

思路

模拟

代码

点击查看代码

#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;

const int mxn = 2e5 + 10;

void solve()
{
	int n, m;
	cin >> n >> m;
	int ans = 0;
	for (int i = 0; i <= m; i++)
	{
		ans += pow(n, i);
		if (ans > 1e9)
		{
			cout << "inf" << endl;
			return;
		}
	}
	cout << ans << 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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C - 2^a b^2

题意

思路

将问题 \(1≤2^a×b^2≤n\) 转化为 \(1\le b \le \sqrt {\frac n {2^a}}\)，枚举 \(a\) 即可

代码

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#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
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 = 0;
	cin >> n;
	for (int i = 1; (1LL << i) <= n; i++)
	{
		int a = (1LL << i);
		int b = sqrtl(n / a);
		ans += ((b + 1) >> 1);
	}
	cout << ans << 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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D - Takahashi the Wall Breaker

题意

思路

当成最短路做，\(dis\) 是前踢的次数
注意前踢的范围是\(2\)(写的时候被卡出💩了)

代码

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#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;

const int mxn = 2e5 + 10;

struct node
{
	int x, y, k;
	node(int x, int y, int k)
	{
		this->x = x;
		this->y = y;
		this->k = k;
	}
	bool operator < (const node& a) const
	{
		return k > a.k;
	}
};

int dx[] = { 1,-1,0,0 };
int dy[] = { 0,0,1,-1 };

void solve()
{
	int h, w, a, b, c, d;
	cin >> h >> w;
	vector<vector<char>> g(h, vector<char>(w));
	for (int i = 0; i < h; i++)
	{
		for (int j = 0; j < w; j++)
		{
			cin >> g[i][j];
		}
	}
	cin >> a >> b >> c >> d;
	a--, b--, c--, d--;
	vector<vector<int>> dis(h, vector<int>(w, LLONG_MAX));
	vector<vector<bool>> vis(h, vector<bool>(w, false));
	priority_queue<node> q;
	dis[a][b] = 0;
	q.push(node(a, b, dis[a][b]));
	while (q.size())
	{
		auto [x, y, k] = q.top();
		q.pop();
		if (vis[x][y]) continue;
		vis[x][y] = true;
		for (int i = 0; i < 4; i++)
		{
			int tx = x + dx[i];
			int ty = y + dy[i];
			if (tx < 0 || ty < 0 || tx >= h || ty >= w) continue;
			if (g[tx][ty] == '.')
			{
				if (dis[tx][ty] > k)
				{
					dis[tx][ty] = k;
					q.push(node(tx, ty, dis[tx][ty]));
				}
			}
			else
			{
				if (dis[tx][ty] > k + 1)
				{
					dis[tx][ty] = k + 1;
					q.push(node(tx, ty, dis[tx][ty]));
				}
				tx += dx[i], ty += dy[i];
				if (tx < 0 || ty < 0 || tx >= h || ty >= w || g[tx][ty] == '.') continue;
				if (dis[tx][ty] > k + 1)
				{
					dis[tx][ty] = k + 1;
					q.push(node(tx, ty, dis[tx][ty]));
				}
			}
		}
	}
	cout << dis[c][d] << 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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E - Ringo's Favorite Numbers 3

题意

思路

由于 \(A\) 是平方数，那数据范围就缩小到了 \(1e6\)，找出范围内每个数的不同质因数个数，并挑出恰有 \(2\) 个质因数的。 要使 \(400数\) 最大，那\(k\)就要最小—— \(2\)。最后只需要找第一个小于 \(A\) 的即可。

代码

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#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;

const int mxn = 1e6 + 10;

vector<int> cnt, num;

void init()
{
	cnt.assign(mxn, 0);
	for (int i = 2; i < mxn; i++)
	{
		if (cnt[i]) continue;
		for (int j = i; j < mxn; j += i)
		{
			cnt[j]++;
		}
	}
	for (int i = 2; i < mxn; i++)
	{
		if (cnt[i] == 2)
		{
			num.push_back(i * i);
		}
	}
}

void solve()
{
	int a;
	cin >> a;
	cout << *prev(upper_bound(num.begin(), num.end(), a)) << endl;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	int __ = 1;
	cin >> __;
	init();
	while (__--)
	{
		solve();
	}
	return 0;
}

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比赛链接 https://atcoder.jp/contests/abc400