11.1 校内模拟赛题解报告

期望得分：100 + 30 + 0 = 130

实际得分：70 + 30 + 0 = 100

总结

考试傻事集合：

T1 用线段树写了个求单调不降的序列的最大值和最小值。

T1 以为 \(mlog^2n\) 能跑过 \(10^6\) 的数据。

T2 \(dp\) 初始值赋错，幸好有大样例 = =；

T1 median

题面

算法一

直接二分中位数，然后用 upper_bound 直接 check 就行。

复杂度 \(O(mlog^2 n)\)

期望得分：\(70pts\)

code

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 7e5 + 5;
const int INF = 1e9 + 7;
int read() {
  int x = 0, f = 1; char c = getchar();
  while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
  while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
  return x * f;
}
int n, m, a[MAXN], Mid, b[MAXN];
int l_1, l_2, r_1, r_2; 
bool Check(int x) {
  int ret_1 = upper_bound(a + l_1, a + r_1 + 1, x) - a;
  int ret_2 = upper_bound(b + l_2, b + r_2 + 1, x) - b;
  ret_1--, ret_2--;
  ret_1 = ret_1 - l_1 + 1, ret_2 = ret_2 - l_2 + 1;
  if(ret_1 + ret_2 >= Mid) return true;
  return false;
}
int tmp[MAXN * 2];
signed main(){
  n = read(), m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) b[i] = read();
  for (int i = 1; i <= m; i++) {
  	 int opt = read();
  	 if(opt == 1) {
  	   int x = read(), y = read(), z = read();
	   if(x == 0) a[y] = z;
  	   else b[y] = z;
	 }
	 else {
	   l_1 = read(), r_1 = read(), l_2 = read(), r_2 = read();
	   int l = min(a[l_1], b[l_2]), r = max(a[r_1], b[r_2]), Ans;
	   Mid = (r_1 - l_1 + 1) + (r_2 - l_2 + 1);
	   Mid = Mid / 2 + 1;
	   while(l <= r) {
	   	 int mid = (l + r) >> 1;
	     if(Check(mid)) Ans = mid, r = mid - 1;
		 else l = mid + 1;  
	   }
	   cout<<Ans<<"\n";
	 }
  }
  return 0;
}

算法二

在 \(a\) 中二分出一个中位数 \(x\)，然后在 \(b\) 中找出它的期望位置 \(pos\)。如果 \(b_{pos - 1} \leq x \leq b_{pos + 1}\) 那么枚举出来的数就是合法的。

注意中位数在 \(a\) 和 \(b\) 中都可能会出现，所以要做两次二分。

这个算法要特判很多情况，所以我就没写。懒

复杂度 \(O(mlogn)\)

期望得分：\(100pts\)

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 5e5 + 10; 
int n,m,t;
int a[MAXN],b[MAXN];
inline int read(){
	int s = 0, f = 0;char ch = getchar();
	while (!isdigit(ch)) f |= ch == '-', ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}
inline int Max(int x,int y) {return x > y ? x : y;}
inline int Min(int x,int y) {return x < y ? x : y;}
inline int Get_Ans_A(int l,int r,int L,int R)
{
  int Len = r - l + 1 + R - L + 1,id = Len / 2;
  int LL = l,RR = r;
  while(LL <= RR) {
    int mid = (LL + RR) >> 1;
    int pos = id - (mid - l + 1) + L;
    if(mid - l == id && a[mid] <= b[L]) return a[mid];
    if(pos > R) {LL = mid + 1;continue;}
    if(pos < L) {RR = mid - 1;continue;}
    if((b[pos + 1] >= a[mid] || pos == R) && a[mid] >= b[pos]) return a[mid];
    if(a[mid] < b[pos]) LL = mid + 1;
    else RR = mid - 1; 
  }
  return 0;
}
inline int Get_Ans_B(int l,int r,int L,int R)
{
  int Len = r - l + 1 + R - L + 1,id = Len / 2;
  int LL = l,RR = r;
  while(LL <= RR) {
    int mid = (LL + RR) >> 1;
    int pos = id - (mid - l + 1) + L;
    if(mid - l == id && b[mid] <= a[L]) return b[mid];
    if(pos > R) {LL = mid + 1;continue;}
    if(pos < L) {RR = mid - 1;continue;}
    if((a[pos + 1] >= b[mid] || pos == R) && b[mid] >= a[pos]) return b[mid];
    if(b[mid] < a[pos]) LL = mid + 1;
    else RR = mid - 1; 
  }
  return 0;
}
signed main(){
  n = read(),m = read();
  for(int i = 1;i <= n;i ++) a[i] = read();
  for(int i = 1;i <= n;i ++) b[i] = read();
  while(m --) {
    int op = read();
    if(op == 1) {
      int x = read(),y = read(),z = read();
      if(x) b[y] = z;
      else a[y] = z;
    }
    else {
      int l = read(),r = read(),L = read(),R = read();
      int Ans = Get_Ans_A(l,r,L,R);
      if(!Ans) Ans = Get_Ans_B(L,R,l,r); 
      printf("%d\n",Ans);
    }
  }
  return 0;
}

算法三

递归求解。

假设当前取得是区间第 \(k\) 大，将 \(k\) 折半，放在两个序列的对应位置上，\(s\) 和 \(t\)，比较一下 \(a_s\) 和 \(b_t\) , 假设 \(a_s < b_t\) ，那么答案就可以变为求区间\([s + 1, r_1], [l_2, r_2]\) 的第 \(\frac {k}{2}\) 大的数。（因为 \(a\) 序列比 \(a[s]\) 小的那些数一定可以全部舍去），然后递归即可。

复杂度 \(O(mlogn)\)

code

/*
递归求解 
*/
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 5e5 + 5;
int read() {
  int x = 0, f = 1; char c = getchar();
  while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
  while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
  return x * f;
}
int n, m;
int a[MAXN], b[MAXN];
int Kth(int ta[], int sta, int tb[], int stb, int k) { 
   if (sta > stb) return Kth(tb, stb, ta, sta, k); // 保证前面那个是短的
   if (sta == 0) return tb[k];  
   if (k == 1) return min(ta[1], tb[1]);
   int ka = min(sta, k / 2), kb = k - ka;
   if (ta[ka] < tb[kb]) return Kth(ta + ka, sta - ka, tb, stb, k - ka);
   return Kth(ta, sta, tb + kb, stb - kb, k - kb);
}
int Query(int la, int ra, int lb, int rb) {
   int sta = ra - la + 1, stb = rb - lb + 1, siz = sta + stb;
   return Kth(a + la - 1, sta,  b + lb - 1, stb, siz / 2 + 1);
}
int main(){
  n = read(), m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) b[i] = read();
  for (int i = 1; i <= m; i++) {
  	 int opt = read();
  	 if(opt == 1) {
  	    int x = read(), y = read(), z = read();
		if(x == 0) a[y] = z;
		else b[y] = z;
	  }
	  else {
	  	int l_1 = read(), r_1 = read(), l_2 = read(), r_2 = read();
	  	printf("%d\n", Query(l_1, r_1, l_2, r_2));
	  }
  }
  return 0;
}

T2 min

题面

算法一：

模拟题意，直接 \(dp\)

\(f_i\) 表示前 \(i\) 个位置，划分得到的最大价值。

转移方程：

\(dp_{i} = \max_{j = 0}^{j = i - 1}\{dp[j] + f(\min_{x = j + 1}^{i} a_x)\}\)

\(dp_0 = 0\)

注意初始化。

复杂度 \(O(n^2)\)

期望得分：30pts

memset(dp, -0x3f3f3f3f3f3f, sizeof dp);
  dp[0] = 0;
  for (int i = 1; i <= n; i++) {
  	for (int j = 0; j < i; j++) {
  	  int x = Query(j + 1, i);
  	  dp[i] = max(dp[i], dp[j] + A * x * x * x + B * x * x + C * x + D);	
	} 
  }
  cout<<dp[n];

算法二

\(A = 0, B = 0, C \leq 0\)

此时 \(f(x) = Cx + D\)

设 \(Cy + D = 0\)

case 1 当 \(z > y\) 时：

\(Cz + D < 0\)

case 2 当 \(w < y\) 时

\(Cw + D > 0\)

对于 case 1 我们可以把它放到 case 2 中的一组中，这样就不会计算 \(case ~1\) 的贡献，因为 \(w < z\)，如果没有 case 2 这种情况，那就只分一组。

\(\sum f(a_i)(f(a_i) > 0)\)

if(A == 0 && B == 0 && C <= 0) {
  	int Ans = 0;
	for (int i = 1; i <= n; i++) {
  	   if(a[i] * C + D >= 0) Ans += a[i] * C + D;	 
	}
	if(Ans > 0) cout<<Ans;
	else {
	  int Min = INF;
	  for (int i = 1; i <= n; i++) Min = min(a[i], Min); 
	  cout<<Min * C + D;
	}
	return 0;
  }

算法三

用单调栈维护 \(g_x = \min_{x = j}^{i}a_x\) 。具体地，单调栈中的元素 \(l_1, l_2\dots l_m\) 表示 \(g_{l_i} \neq g_{l_i - 1}\) 的每个 \(l_i\)（就是最小值变化的转折
点），那么有 \(\forall x \in[l_i, l_{i + 1} - 1], g_x\) 相同，此时 \(dp\) 值最大的那个点一定最优秀，于是维护 \(h_i = \max_{x = l_i}^{l_i + 1}{dp[x]}\)，表示每个取最小值元素对应区间的最优答案。

这样的话，每一次的答案就是 \(\max\{h_i + f(g_{l_i})\}\)，采用一棵线段树
或者可删除堆维护单调栈即可。

复杂度 \(O(nlogn)\)

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MAXN = 524287;
const long long inf = 0x3f3f3f3f3f3f3f3f;
int read() {
  int x = 0, f = 1; char c = getchar();
  while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
  while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
  return x * f;
}
int a[MAXN], st[MAXN], tp, n, A, B, C, D;
ll T[MAXN << 1], f[MAXN], mx[MAXN];
void Up(int i, ll x) {
  for (T[i += MAXN] = x; i >>= 1;) T[i] = max(T[i << 1], T[i << 1 | 1]); 
}
ll Cal(ll x) {return ((A * x + B) * x + C) * x + D; }
int main(){
  n = read(), A = read(), B = read(), C = read(), D = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  memset(T, -0x3f, sizeof T);
  f[0] = 0, mx[1] = 0, st[tp = 1] = a[1];
  Up(1, Cal(a[1]));
  for (int i = 1; i <= n; i++) {
  	 f[i] = T[1]; ll x = f[i];
  	 while(st[tp] > a[i + 1] && tp) x = max(x, mx[tp]), Up(tp--, -inf);
	 st[++tp] = a[i + 1], mx[tp] = x, Up(tp, x + Cal(st[tp]));
  }
  cout<<f[n];
  return 0;
}

T3

题面

std

#include<cstdio>
#include<algorithm>
#define ls p << 1
#define rs p << 1 | 1
#define rt 1, 1, Q
const int N = 1e5 + 10, Y = 2e5 + 10, P = 1e9 + 7;
int ri() {
	char c = getchar(); int x = 0, f = 1; for(;c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
	for(;c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) - '0' + c; return x * f;
}
int t[N << 2], tm[N << 2], c[Y], b[Y], l[Y], r[Y], pr[Y], to[Y << 1], nx[Y << 1], Q, n, m, k, H;
void add(int x, int p) {to[++H] = 1LL * to[pr[x]] * (1 - p) % P; nx[H] = pr[x]; pr[x] = H;}
struct D {int l, r;} q[N];
struct X {int x, y, p;}p[Y];
bool cmp1(X a, X b) {return a.y < b.y;}
bool cmp2(D a, D b) {return a.l == b.l ? a.r < b.r : a.l < b.l;}
int Pow(int x, int k) {
	int r = 1;
	for(;k;x = 1LL * x * x % P, k >>= 1) if(k & 1) r = 1LL * r * x % P;
	return r;
}
void B(int p, int L, int R) {
	tm[p] = 1; if(L == R) return void(t[p] = 1);
	int m = L + R >> 1; B(ls, L, m); B(rs, m + 1, R); 
	t[p] = t[ls] + t[rs];
}
void Tag(int p, int v) {tm[p] = 1LL * tm[p] * v % P; t[p] = 1LL * t[p] * v % P;}
void Pd(int p) {if(tm[p] != 1) Tag(ls, tm[p]), Tag(rs, tm[p]), tm[p] = 1;}
void M(int p, int L, int R, int st, int ed, int v) {
	if(L == st && ed == R) return Tag(p, v);
	int m = L + R >> 1; Pd(p);
	if(st <= m) M(ls, L, m, st, std::min(ed, m), v);
	if(ed > m) M(rs, m + 1, R, std::max(st, m + 1), ed, v);
	t[p] = (t[ls] + t[rs]) % P;
}
void C(int x) {
	if(l[x] > r[x]) return ; 
	int m = Pow(1 - (to[pr[x]] - b[x]) % P, P - 2); pr[x] = nx[pr[x]]; 
	M(rt, l[x], r[x], 1LL * (1 - (to[pr[x]] - b[x]) % P) * m % P);
}
int main() {
	//freopen("max.in","r",stdin);
	//freopen("max.out","w",stdout);
	n = ri(); m = ri(); Q = ri(); int tp = 0;
	for(int i = 1;i <= m; ++i) {
		int x = ri(), y = ri(), px = ri();
		if(!px || !y) continue;
		p[++tp].x = x; p[tp].y = y; p[tp].p = px;
	}
	std::sort(p + 1, p + tp + 1, cmp1);
	for(int i = 1;i <= n; ++i) to[++H] = 1, pr[i] = H;
	for(int i = 1;i <= tp; ++i) add(p[i].x, p[i].p);
	for(int i = 1;i <= n; ++i) b[i] = to[pr[i]];
	for(int i = 1;i <= Q; ++i) q[i].l = ri(), q[i].r = ri();
	std::sort(q + 1, q + Q + 1, cmp2);
	int L = 1, R = 0; 
    for(int i = 1;i <= n; ++i) {
        for(;L <= R && q[L].r < i; ++L) ;
        for(;q[R + 1].l <= i && R < Q; ++R) ;
        l[i] = L; r[i] = R;
    } 
    B(rt); int A = 0; p[0].y = 0;
    for(int i = tp, j; i; i = j) {
    	for(j = i;p[j].y == p[i].y && j; --j) C(p[j].x);
    	A = (A + 1LL * t[1] * (p[i].y - p[j].y)) % P;
    }
    A = (1LL * p[tp].y * Q - A) % P;
    printf("%d\n", (A + P) % P);
	return 0;
}

posted @ 2021-11-01 20:04 Dita 阅读(64) 评论(0) 收藏举报

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11.1 校内模拟赛题解报告

T1 median

T2 min

T3

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