Codeforces Round #287 (Div. 2) A、B、C、D、E

标签：

A：签到题，排序判断一下能学几门即可

B：圆心可以每步可以移动2 * r的距离，方向任选，所以答案是ceil(两点距离 / 2 / r)

C：递归下去就可以了，dfs(h, n, flag)，h表示当前到哪层，n表示当前层下的出口相对位置，flag表示下一步往左还是往右

D：数位DP，从最低位往最高位去放数字，如果一旦出现取模为0，就可以直接计算种数位后面还剩多少位，最后一位可以放1-9，其他都是0-9，不然就继续记忆化搜下去

E：最短路，把图建起来，假设1边总数为x，选择的路径上1边数位a，0边数为b，那么答案就是x - a + b条，答案要尽量小，所以最短路选1的优先级要大于选0的优先级，这个处理方式可以直接把边权扩大10W倍，然后0边多+1即可

代码：

A：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 105;

int n, k;

struct SB {
	int a, id;
} sb[N];

bool cmp(SB a, SB b) {
	return a.a < b.a;
}

int main() {

	scanf("%d%d", &n, &k);
	for (int i = 0; i < n; i++) {
		scanf("%d", &sb[i].a);
		sb[i].id = i;
	}
	sort(sb, sb + n, cmp);
	int tot = 0;
	int cnt = 0;
	for (int i = 0; i < n; i++) {
		if (tot + sb[i].a > k) break;
		tot += sb[i].a;
		cnt++;
	}
	printf("%d\n", cnt);
	for (int i = 0; i < cnt; i++)
		printf("%d ", sb[i].id + 1);
	return 0;
}

#include <cstdio>
#include <cstring>
#include <cmath>

double r, xa, ya, xb, yb;
const double eps = 1e-8;

int main() {
	scanf("%lf%lf%lf%lf%lf", &r, &xa, &ya, &xb, &yb);
	double d = sqrt((xb - xa) * (xb - xa) + (yb - ya) * (yb - ya));
	printf("%.0lf\n", ceil(d / r / 2));
	return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;

int h;
ll n;

ll dfs(int d, ll n, int lr) {
	if (d == h) return 0;
	if (!lr && n > (1LL<<(h - d - 1)))
		return (1LL<<(h - d)) + dfs(d + 1, n - (1LL<<(h - d - 1)), 0);
	if (lr && n <= (1LL<<(h - d - 1)))
		return (1LL<<(h - d)) + dfs(d + 1, n, 1);
	if (!lr && n <= (1LL<<(h - d - 1)))
		return dfs(d + 1, n, 1) + 1;
	if (lr && n > (1LL<<(h - d - 1)))
		return dfs(d + 1, n - (1LL<<(h - d - 1)), 0) + 1;
}

int main() {
	scanf("%d%lld", &h, &n);
	printf("%lld\n", dfs(0, n, 0));
	return 0;
}

#include <cstdio>
#include <cstring>

typedef long long ll;
const int N = 1005;
const int M = 105;

int n, k;
ll m, pow10[N];
ll dp[N][M][2];

ll pow_mod(ll x, int k) {
	ll ans = 1;
	while (k) {
		if (k&1) ans = ans * x % m;
		x = x * x % m;
		k >>= 1;
	}
	return ans;
}

ll dfs(int u, int mod, int flag) {
	ll &ans = dp[u][mod][flag];
	if (ans != -1) return ans;
	ans = 0;
	if (u == n) return ans;
	int s = 0;
	if (u == n - 1) s = 1;
	for (int i = s; i <= 9; i++) {
		int next = (pow10[u] * i + mod) % k;
		int f = flag;
		if (i > 0) f = 1;
		if (next == 0 && f) {
			if (n - u - 2 >= 0)
				ans = (ans + pow_mod(10LL, n - u - 2) * 9 % m) % m;
			else ans = (ans + 1) % m;
		}
		else {
			ans = (ans + dfs(u + 1, next, f)) % m;
		}
	}
	return ans;
}

int main() {
	scanf("%d%d%lld", &n, &k, &m);
	pow10[0] = 1;
	for (int i = 1; i < N; i++)
		pow10[i] = pow10[i - 1] * 10 % k;
	memset(dp, -1, sizeof(dp));
	printf("%lld\n", dfs(0, 0, 0));
	return 0;
}

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

#define INF 0x3f3f3f3f3f3f3f

typedef long long ll;
const int MAXNODE = 100005;
const int N = 100005;

struct Edge {
	int u, v, use;
	ll dist;
	Edge() {}
	Edge(int u, int v, ll dist) {
		this->u = u;
		this->v = v;
		this->dist = dist;
		this->use = 0;
	}
};

Edge out[N];
int on;

struct HeapNode {
	ll d;
	int u;
	HeapNode() {}
	HeapNode(ll d, int u) {
		this->d = d;
		this->u = u;
	}
	bool operator < (const HeapNode& c) const {
		return d > c.d;
	}
};

struct Dijkstra {
	int n, m;
	vector<Edge> edges;
	vector<int> g[MAXNODE];
	bool done[MAXNODE];
	ll d[MAXNODE];
	int p[MAXNODE];

	void init(int tot) {
		n = tot;
		for (int i = 1; i <= n; i++)
			g[i].clear();
		edges.clear();
	}

	void add_Edge(int u, int v, ll dist) {
		edges.push_back(Edge(u, v, dist));
		m = edges.size();
		g[u].push_back(m - 1);
	}

	void dijkstra(int s) {
		priority_queue<HeapNode> Q;
		for (int i = 1; i <= n; i++) d[i] = INF;
		d[s] = 0;
		p[s] = -1;
		memset(done, false, sizeof(done));
		Q.push(HeapNode(0, s));
		while (!Q.empty()) {
			HeapNode x = Q.top(); Q.pop();
			int u = x.u;
			if (done[u]) continue;
			done[u] = true;
			for (int i = 0; i < g[u].size(); i++) {
				Edge& e = edges[g[u][i]];
				if (d[e.v] > d[u] + e.dist) {
					d[e.v] = d[u] + e.dist;
					p[e.v] = g[u][i];
					Q.push(HeapNode(d[e.v], e.v));
				}
			}
		}
	}
	
	void print() {
		int u = n;
		on = 0;
		while (p[u] != -1) {
			edges[p[u]].use = 1;
			edges[p[u]^1].use = 1;
			u = edges[p[u]].u;
		}
		for (int i = 0; i < edges.size(); i += 2) {
			if (edges[i].use == 0 && edges[i].dist == 100000) {
				out[on] = edges[i];
				out[on].dist = 0;
				on++;
			} else if (edges[i].use == 1 && edges[i].dist == 100001) {
				out[on] = edges[i];
				out[on].dist = 1;
				on++;
			}
		}
		printf("%d\n", on);
		for (int i = 0; i < on; i++) {
			int a = out[i].u;
			int b = out[i].v;
			ll c = out[i].dist;
			printf("%d %d %lld\n", a, b, c);
		}
	}
} gao;

int n, m;


int main() {
	scanf("%d%d", &n, &m);
	gao.init(n);
	int u, v, w;
	while (m--) {
		scanf("%d%d%d", &u, &v, &w);
		if (w == 1) {
			gao.add_Edge(u, v, 100000);
			gao.add_Edge(v, u, 100000);
		} else {
			gao.add_Edge(u, v, 100001);
			gao.add_Edge(v, u, 100001);
		}
	}
	gao.dijkstra(1);
	gao.print();
	return 0;
}

Codeforces Round #287 (Div. 2) A、B、C、D、E

标签：

原文地址：http://blog.csdn.net/accelerator_/article/details/43079273