51nod 模拟2

本文主要是介绍51nod 模拟2，对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

A. 直接pow,代码略
B
分子分母分开处理
\(a/b\)转移到\(\frac{\frac{a}{b}+\frac{b}{a}}{2} = \frac{a^2+b^2}{2ab}\)
然后\(a'=a^2+b^2, b'=2ab\)所以\(a'+b'=(a+b)^2, a'-b' = (a-b)^2\)
可以找规律完成递推

%:pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
const int L = 1E5+10, R = 500000002;
#define int ll
typedef long long ll;
const ll MOD = 1e9+7, PHI = MOD-1;
ll qpow(ll a, ll b,ll mod){
	ll ret = 1;
	for(;b;b>>=1){
		if(b & 1) ret = ret * a %mod;
		a = a * a % mod;
	}
	return ret;
}
ll inv(ll x){
	return qpow(x, MOD-2, MOD);
}
ll get(ll a, ll b){
	if(b == 0) return a;
	ll x = (a + 1) * inv(2) % MOD, y = (a - 1) * inv(2) % MOD;
	ll p = qpow(x, (qpow(2, b, PHI)), MOD) , q = qpow(y, qpow(2, b, PHI),MOD);
	return (p + q)%MOD * inv(p - q)%MOD;
}
void solve(){
	ll x, n;
	cin >> x >> n;
	--n;
	cout << (get(x, n)+ MOD)%MOD << endl;
}
signed main(){
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	int T;
	cin >> T;
	while(T--){
		solve();
	}
}

C
非常有趣
我们的目的是把每个点的最小alpha求出来
如何求？一条边有两个方向，一条边和一个方向可以决定一个贡献
我们把有向边定向，然后求每条有向边的alpha
每条有向边\((u,v)\)可以对\(u\)的子树产生贡献
对子树贡献可以用线段树区间取\(max\)维护
线段树每个叶子\(u\)表示\(u\)为根时的最小alpha

#include<bits/stdc++.h>
using namespace std;
typedef long double ld;
const int N = 1e5 + 10, TR = 4 * N, E = 2 * N;
const ld eps = 1e-14;
int n,k, q;
struct segment_tree{
	struct node{
		ld laz;
	}tree[TR];
	void update(int rt, ld val){
		tree[rt].laz = max(tree[rt].laz, val);
	}
	void modify(int rt,int l, int r, int s, int t, ld val){
		if(s > t) return;
		if(s <= l && r <= t){
			tree[rt].laz = max(tree[rt].laz, val);
			return;
		}
		int mid = (l + r) >> 1;
		if(s <= mid) modify(rt<<1, l, mid, s ,t, val);
		if(t > mid) modify(rt<<1|1, mid+ 1, r, s ,t, val);
	}
	ld query(int rt, int l, int r, int s){
		ld ret = tree[rt].laz;
		if(l >= r){
			return ret;
		}
		int mid = (l + r) >> 1;
		if(s <= mid)
			return max(ret, query(rt<<1, l, mid, s));
		else 
			return max(ret, query(rt<<1|1, mid + 1, r, s));
	}
}segt;
struct graph{
	struct edge{
		int nxt, to;
	}e[E];
	graph(){
		elen = 1;
	}
	int elen, head[N];
	void inse(int frm, int to){
		e[++elen] = {head[frm], to};
		head[frm] = elen;
		++deg[to];
	}
	void insf(int u, int v){
		inse(u, v),inse(v, u);
	}
	int fat[N];
	int siz[N], deg[N];
	int son1[N], son2[N];
	int dfn[N], tim;
	ld val[E];
	void dfs1(int u){
		siz[u] = 1;
		for(int i = head[u]; i; i = e[i].nxt){
			int v = e[i].to;
			if(v == fat[u]) continue;
			fat[v] = u;
			dfs1(v);
			siz[u] += siz[v];
			if(siz[v] > siz[son1[u]]){
				son2[u] = son1[u], son1[u] = v;
			}else if(siz[v] > siz[son2[u]]){
				son2[u] = v;
			}
		}
	}
	void dfs2(int u){
		for(int i = head[u]; i;i = e[i].nxt){
			int v = e[i].to;
			if(v == fat[u]) continue;
			val[i] = deg[v] <= 2 ? 0 : ld(siz[son1[v]]) /siz[v];
			dfs2(v);
		}
		for(int j  = head[u]; j;j=  e[j].nxt){
			int i = (j ^ 1);
			int v = e[j].to;
			if(v == fat[u]) continue;
			int mxp = (v == son1[u]) ? son2[u] : son1[u];
			int mxs = max( n - siz[u], siz[mxp]);
			val[i] = deg[u] <= 2 ? 0 :ld(mxs) / (n - siz[v]);
		}
	}
	void dfs3(int u){
		dfn[u] = ++tim;
		for(int i = head[u]; i; i=  e[i].nxt){
			int v = e[i].to;
			if(v == fat[u]) continue;
			dfs3(v);
		}
	}
	void dfs4(int u){
		for(int i = head[u]; i;i = e[i].nxt){
			int v = e[i].to;
			if(v == fat[u]) continue;
			int j = i ^ 1;
			segt.modify(1, 1, n, dfn[v], dfn[v] + siz[v] - 1, val[j]);
			dfs4(v);
		}
		for(int i = head[u]; i;i = e[i].nxt){
			int v = e[i].to;
			if(v == fat[u]) continue;
			segt.modify(1, 1, n, 1, dfn[v] - 1, val[i]);
			segt.modify(1, 1 ,n, dfn[v] + siz[v], n ,val[i]);
		}
	}
}G;
ld wei[N];
int main(){
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> k;
	for(int i = 1; i < n; ++i){
		int u, v;
		cin >> u >> v;
		G.insf(u, v);
	}
	G.dfs1(1);
	G.dfs2(1);
	G.dfs3(1);
	G.dfs4(1);
	for(int i = 1; i <= n; ++i){
		wei[i] = segt.query(1, 1, n, G.dfn[i]);
	}
	for(int i = 1; i <= n; ++i){
		if(G.deg[i] >= 2){
			int msiz = max(G.siz[G.son1[i]], n - G.siz[i]);
			wei[i] = max(wei[i], (ld)1.0 * msiz / n);
		}
	}
	sort(wei + 1, wei + 1 + n);
	cin >> q;
	int las = 0;
	for(int i = 1; i <= q ; ++i){
		int u , v;
		cin >> u >> v;
		u = u xor (las * k), v = v xor(las * k);
		ld key = ld(u)/v;
		las = (upper_bound(wei + 1, wei + 1 + n, key + eps) - wei - 1);
		cout << las << endl;
	}
	return 0;
}

D
发现价钱和天数是等价的，于是我们设工厂的两个关建值为\(a, b\)
商店为\(c,d\)
\(a,b\)越小越好，\(c,d\)越大越好
我们分别排序，并且去掉一定不可能产生贡献的点（若\(a, b\)均比另一个点大）
接下来发现有决策单调性，分治即可
证明：设\(u\)为商店，\(i,j\)为两个工厂\(i<j\),若\(j\)比\(i\)优，则
\((c_u - a_i) * (d_u - b_i) < (c_u - a_j) * (d_u - b_j)\)
展开后移项，得
\(d_u(a_j-a_i) + a_jb_j< c_u(b_i-b_j) + a_kb_k\)
因为\(i<j\)，所以\(a_i-a_i > 0\) ,\(b_i - b_j>0\)
又因为对\(v > u\), \(d_v < d_u, c_v < c_u\)所以\(j仍然比\)i$优

#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5e5+10;
struct t_a{
	int a, b;
	bool operator < (const t_a &rhs) const{
		return tie(a, b) < tie(rhs.a, rhs.b);
	}
}x[N],t1[N];
struct t_b{
	int c, d;
	bool operator <(const t_b & rhs)const{
		return tie(d, c) > tie(rhs.d, rhs.c);
	}
}y[N],t2[N];
ll ans = 0;
ll calc(t_a a, t_b b){
	if(b.c - a.a < 0 || b.d - a.b < 0) return 0;
	return 1ll * (b.c - a.a) * (b.d- a.b);
}
void solve(int l, int r, int s, int t){
	if(l > r || s > t)  return;
	int mid = (l + r) >> 1;
	int pos = s;
	for(int i = s + 1; i <= t; ++i){
		if(calc(t1[i], t2[mid]) >= calc(t1[pos], t2[mid])){
			pos = i;
		}
	}
	ans = max(ans, calc(t1[pos], t2[mid]));
	solve(l, mid - 1, s, pos);
	solve(mid + 1, r, pos, t);
}
int m, n;
int main(){
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> m >> n;
	for(int i = 1; i <= m; ++i){
		cin >> x[i].a >> x[i].b;
	}
	for(int i = 1; i <= n; ++i){
		cin >> y[i].c >> y[i].d;
	}
	sort(x + 1, x + m + 1);
	sort(y + 1 ,y + n + 1);
	int c1 = 0, c2 = 0;
	t1[++c1] = x[1];
	t2[++c2] = y[1];
	for(int i = 2; i <= m; ++i){
		if(t1[c1].b <= x[i].b) continue;
		t1[++c1] = x[i];
	}
	for(int i = 2; i <= n; ++i){
		if(t2[c2].c >= y[i].c) continue;
		t2[++c2] = y[i];
	}
	m = c1, n = c2;
	for(int i = 1; i <= n; ++i){
		int pos = lower_bound(t1 + 1, t1 + 1 + m, t_a{t2[i].c, -1}) - t1 - 1;
		if(pos == 0 || t1[pos].b >= t2[i].d){
			t2[i].c = t2[i].d = -1;
		}
	}
	c2 = 0;
	for(int i = 1; i <= n; ++i){
		if(t2[i].c == -1) continue;
		y[++c2] = t2[i];
	}
	n = c2;	
	memcpy(t2, y, sizeof(t2));
	solve(1, n, 1, m);
	cout << ans << endl;
	return 0;
}

这篇关于51nod 模拟2的文章就介绍到这儿，希望我们推荐的文章对大家有所帮助，也希望大家多多支持为之网！