AtCoder Beginner Contest 132

[TOC]

Contest Info

---

[Practice Link](https://atcoder.jp/contests/abc132/tasks)

Solved	A	B	C	D	E	F
6/6	O	O	O	O	Ø	O

• O 在比赛中通过
• Ø 赛后通过
• ! 尝试了但是失败了
• - 没有尝试

Solutions

---

A. Fifty-Fifty

#include <bits/stdc++.h>
using namespace std;

int main() {
	string s; cin >> s;
	sort(s.begin(), s.end());
	s.erase(unique(s.begin(), s.end()), s.end());
	cout << (((int)s.size() == 2) ? "Yes" : "No") << "
";
	return 0;
}

B. Ordinary Number

#include <bits/stdc++.h>
using namespace std;

#define N 50
int n, a[N];

int main() {
	while (scanf("%d", &n) != EOF) {
		for (int i = 1; i <= n; ++i) {
			scanf("%d", a + i);
		}
		int res = 0;
		for (int i = 2; i < n; ++i) {
			int f = a[i] < a[i - 1];
			int g = a[i] < a[i + 1];
			if (f ^ g) {
				++res;
			}
		}
		printf("%d
", res);
	}
	return 0;
}

C. Divide the Problems

#include <bits/stdc++.h>
using namespace std;

#define N 100010
int n, a[N];

int main() {
	while (scanf("%d", &n) != EOF) {
		for (int i = 1; i <= n; ++i) {
			scanf("%d", a + i);
		}
		sort(a + 1, a + 1 + n);
		int mid = n / 2;
		if (a[mid] == a[mid + 1]) {
			puts("0");
		} else {
			printf("%d
", a[mid + 1] - a[mid]);
		}
	}
	return 0;
}

D. Blue and Red Balls

题意： 有$K$个蓝球，$N - K$个红球，询问将蓝球分成$i(1 leq i leq k)$堆，中间用红球隔开的方案数分别是多少?

思路： 考虑先将$k$个蓝球分成$i$堆，然后每堆后面跟着一个红球，然后就有$i + 1$个空隙，剩下的红球相当于要放入$i + 1$个空箱中，允许空箱的经典问题。

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 2010
const int p = 1e9 + 7;
int n, k;
ll fac[N], inv[N];
ll qmod(ll base, ll n) {
	ll res = 1;
	while (n) {
		if (n & 1) {
			res = res * base % p;
		}
		base = base * base % p;
		n >>= 1;
	}
	return res;
}

ll C(int n, int m) {
	if (n < m) return 0;
	return fac[n] * inv[m] % p * inv[n - m] % p;
}

ll f(int n, int i, int k) {
	ll remind = n - k;
	if (n - k < i - 1) {
		return 0;
	}
	remind -= i - 1;
	return C(k - 1, i - 1) * C(i + remind, i) % p; 
}

int main() {
	fac[0] = 1;
	for (int i = 1; i <= 2000; ++i) {
		fac[i] = fac[i - 1] * i % p;
	}
	inv[2000] = qmod(fac[2000], p - 2);
	for (int i = 2000; i >= 0; --i) {
		inv[i - 1] = inv[i] * i % p;
	}
	while (scanf("%d%d", &n, &k) != EOF) {
		for (int i = 1; i <= k; ++i) {
			printf("%lld
", f(n, i, k));
		}
	}
	return 0;
}

E. Hopscotch Addict

题意： 求$S$到$T$的最短路，并且要满足其长度是$3$的倍数。

思路： $dis[i][j]$表示到第$i$个点，长度模$3$的值为$j$的最短路是多少，然后跑Dijkstra即可。

#include <bits/stdc++.h>
using namespace std;

#define INF 0x3f3f3f3f
#define N 100010
int n, m, s, t;
vector <vector<int>> G;
struct node {
	int u, w;
	node() {}
	node (int u, int w) : u(u), w(w) {}
	bool operator < (const node &other) const {
		return w > other.w;
	}
};

int dis[N][3], used[N][3];
void BFS() {
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j < 3; ++j) {
			dis[i][j] = INF;
			used[i][j] = 0;		
		}
	}
	dis[s][0] = 0;	
	priority_queue <node> pq;
	pq.push(node(s, 0)); 
	while (!pq.empty()) {
		int u = pq.top().u, w = pq.top().w; pq.pop(); 
		if (used[u][w % 3]) continue;
		used[u][w % 3] = 1;
		for (auto v : G[u]) {
			if (dis[v][(w + 1) % 3] > w + 1) {
				dis[v][(w + 1) % 3] = w + 1;  
				pq.push(node(v, w + 1));  
			}
		}
	}
}

int main() {
	while (scanf("%d%d", &n, &m) != EOF) {
		G.clear(); G.resize(n + 1);
		for (int i = 1, u, v; i <= m; ++i) {
			scanf("%d%d", &u, &v);
			G[u].push_back(v);
		}
		scanf("%d%d", &s, &t);
		BFS();
		if (dis[t][0] == INF) puts("-1");
		else {
			printf("%d
", dis[t][0] / 3);
		}
	}
	return 0;
}

F. Small Products

题意： 构造一个长度为$k$的，并且相邻两数之积不超过$N$的序列的方案数是多少？

思路： $f[i][j]$表示到第$i$位，当前位为$j$的方案数是多少。 暴力$DP$显然不行。 但是我们考虑我们每次对于$j$转移的都是$sumlimits_^{N / j} f[i - 1][j]$，考虑到$N / j$的取值只有$2sqrt$种，所以只需要存$k cdot 2sqrt$种状态。转移即可。

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 110
#define M 100010
#define pii pair <int, int>
#define fi first
#define se second
const ll p = 1e9 + 7;
int n, k;
ll f[N][M];
pii g[M]; 
map <int, int> mp;

int main() {
	while (scanf("%d%d", &k, &n) != EOF) {
		memset(f, 0, sizeof f);
		mp.clear();
		int m = 0;
		for (int i = 1, j; i <= k; i = j + 1) {
			j = k / (k / i);
			g[++m] = pii(j, j - i + 1);
			mp[j] = m;
		}
		for (int i = 1; i <= m; ++i) {
			f[1][i] = (f[1][i - 1] + g[i].se) % p; 
		}
		for (int i = 2; i <= n; ++i) {
			for (int j = 1; j <= m; ++j) {
				f[i][j] = (f[i][j - 1] + 1ll * g[j].se * f[i - 1][mp[k / g[j].fi]] % p) % p;
			}
		}
		printf("%lld
", f[n][m]);
	}
	return 0;
}