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Codeforces Round #266 (Div. 2)

题目链接

A：就简单的判断一下那种更大即可

B：枚举x到sqrt(n)，然后可以直接算出y，然后判断一下即可

C：先判断和是否是3的倍数，然后预处理出前缀和出现位置和后缀和对应sum / 3个数，然后从头往后扫一遍把当前一个和后面进行组合即可

D：先预处理出差分，使得数组表示线段的添加方式，然后每次有一个-1，就能和前面多少个1进行匹配，方案数就乘上多少，如果是0，就能和前面+1个匹配

E：利用并查集，把每次询问拆分成2个部分，起点到x，x到根，然后每次从根往下dfs一遍，对应询问符合的就把对应询问++，dfs完如果一个询问符合两次，就是符合的输出YES，否则就是NO

代码：

#include <cstdio>
#include <cstring>

int n, m, a, b;

int solve() {
	if (b >= m * a) return a * n;
	int yu = n % m;
	int ans = n / m * b;
	if (yu * a < b) return ans + yu * a;
	return ans + b;
}

int main() {
	scanf("%d%d%d%d", &n, &m, &a, &b);
	printf("%d\n", solve());
	return 0;
}

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

typedef long long ll;

ll n, a, b;

int main() {
	scanf("%lld%lld%lld", &n, &a, &b);
	n = n * 6;
	ll ans = 1e18, x, y;
	if (a * b >= n) {
		x = a;
		y = b;
		ans = a * b;
	}
	else {
		int flag = 0;
		if (a > b) {
			flag = 1;
			swap(a, b);
		}
		for (int i = 1; i < 1000000 && i < n; i++) {
			ll r = n / i + (n % i != 0);
			ll l = i;
			if (l > r) swap(l, r);
			if (l < a || r < b) continue;
			if (i * r < ans) {
				ans = i * r;
				x = i;
				y = r;
			}
		}
		if (flag) swap(x, y);
	}
	printf("%lld\n%lld %lld\n", ans, x, y);
	return 0;
}

#include <cstdio>
#include <cstring>

const int N = 500005;

typedef long long ll;

int n;
ll a[N], pres[N], prec[N], sufs[N], sufc[N];

int main() {
	scanf("%d", &n);
	ll sum = 0;
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &a[i]);
		sum += a[i];
	}
	if (sum % 3) printf("0\n");
	else {
		sum /= 3;
		for (int i = 1; i <= n; i++) {
			pres[i] = pres[i - 1] + a[i];
			if (pres[i] == sum)
				prec[i] = 1;
		}
		for (int i = n; i >= 1; i--) {
			sufs[i] = sufs[i + 1] + a[i];
			sufc[i] = sufc[i + 1];
			if (sufs[i] == sum) sufc[i]++;
		}
		ll ans = 0;
		for (int i = 1; i <= n; i++)
			ans += prec[i] * sufc[i + 2];
		printf("%lld\n", ans);
	}
	return 0;
}

#include <cstdio>
#include <cstring>

typedef long long ll;
const int MOD = 1000000007;

const int N = 2005;
int n, h, a[N], b[N];

int main() {
	scanf("%d%d", &n, &h);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i] = h - a[i];
	for (int i = 1; i <= n + 1; i++) b[i] = a[i] - a[i -1];
	int ans = 1, cnt = 0;
	for (int i = 1; i <= n + 1; i++) {
		if (b[i] == 0) ans = (ll)ans * (cnt + 1) % MOD;
		else if (b[i] == 1) cnt++;
		else if (b[i] == -1) ans = (ll)ans * cnt % MOD, cnt--;
		else ans = 0;
	}
	printf("%d\n", ans);
	return 0;
}

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

#define MP(a,b) make_pair(a,b)
typedef pair<int, int> pii;
const int N = 100005;

int n, m, parent[N];

int find(int x) {
	return x == parent[x] ? x : parent[x]  = find(parent[x]);
}

vector<pii> p, q[N];
vector<int> g[N];
int tot, vis[N], cnt[N];

void dfs(int u) {
	vis[u] = 1;
	for (int i = 0; i < g[u].size(); i++)
		dfs(g[u][i]);
	for (int i = 0; i < q[u].size(); i++) {
		if (vis[q[u][i].first])
			cnt[q[u][i].second]++;
	}
	vis[u] = 0;
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)
		parent[i] = i;
	int c, x, y;
	while (m--) {
		scanf("%d%d", &c, &x);
		if (c == 2)
			p.push_back(MP(find(x), x));
		else {
			scanf("%d", &y);
			if (c == 1) {
				g[y].push_back(x);
				int px = find(x);
				int py = find(y);
				if (px != py)
					parent[px] = py;
			} else {
				q[x].push_back(MP(p[y - 1].first, tot));
				q[p[y - 1].second].push_back(MP(x, tot));
				tot++;
			}
		}
	}
	for (int i = 1; i <= n; i++)
		if (parent[i] == i) dfs(i);
	for (int i = 0; i < tot; i++)
		if (cnt[i] == 2) printf("YES\n");
		else printf("NO\n");
	return 0;
}

Codeforces Round #266 (Div. 2)

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原文地址：http://blog.csdn.net/accelerator_/article/details/39522585