标签：最优 解题报告 端点 isp val mat read main efi
期望得分：\(30 + 30 + 0\)
实际得分：\(30 + 0 + 0\)
炸了，这次考试完全炸了，\(T1\)只会打暴力，\(T2\)神奇大\(DP\)，\(T3\)概率期望迷
\(T1\)想不出\(70\)分来，只能默默\(orz\ cgp\)大佬
\(T2\)打错了……\(awsl\)
\(T3\)是个鬼……\(毒\)不懂
先打\(30\)分暴力，这个很好打
然后\(70\)分\(O(n\log^2)\)怎么写呢
考虑二分答案
二分的左端点为\(0\)，右端点为整整两个序列中的最大值\(maxn\)，\(len\)先计算出取出的区间长度的中间值，然后进行二分，用\(upper\_bound\)计算出\(A\)序列中在\(mid\)左边的数的个数\(now1\)，以及\(B\)序列中在\(mid\)左边的数的个数\(now2\)，若他们两个的长度大于\(len\)就缩小范围到\(mid\)左边，相应的缩小\(r1\)和\(r2\)的范围，反之则缩小范围到\(mid\)右边，相应扩大\(l1\)和\(l2\)，最后输出\(ans\)就好了
满分咋做？(\(from\ solution\))
实际上我们可以进行一波分类讨论。可以通过当前数的位置得到
这个数在另一个序列的期望位置。假设当前的数为\(x\)，期望位置的数为 \(y\)，下一个数为\(z\)，那么 \(z\le x \le y\)时\(x\)就是答案，否则比较一下大小，往两边跳。
这种方法要特判很多种情况。事实上，我们还有一种较为简便，
普适性更强的方法。假设当前要取的是区间的第\(k\)大，将\(k\)折半，放在两个区间的对应位置 \(s, t\)上，比较 \(a[s], b[t]\)，不妨设\(a[s] < b[t]\)，那么答案可以化归至区间\([l1, s - 1],[l2, r2]\)的第\(\frac{k}{2}\)大数（因为\(a\)序列比\(a[s]\)小的那些数一定可以全部舍去）， 递归即可
\(30\)分的话，可以想到一个\(n^2\)的\(dp\)方程
用\(dp[i]\)表示分割\([l,i]\)的最大答案，转移方程为
\[dp[i] = \max_{j = 0}^{i - 1}dp[j] + f(\min_{x = j + 1}^{i} a_x)，dp[0] = 0\]
满分做法？(同样\(from\ solution\))
显然可以采用一个单调递增的栈来维护\(g_x = \min_{x = j}^{i}a[x]\)具体地，单调栈中的元素\(l_1,l_2,…l_m\)表示\(g_{l_i}!=g_{l_{i-1}}\)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\(h_i = \max_{x = l_i}^{l_i +1}dp[x]\)，表示每个取到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。
不会，咕咕咕
考场三十分代码
/*
By:Loceaner
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
return x * f;
}
const int N = 5e5 + 11;
int n, m, a[N], b[N], now[N], cnt, tot, ans[N];
int opt, x, y, z, l1, l2, r1, r2;
void solve() {
while(m--) {
opt = read();
if(opt == 1) {
x = read(), y = read(), z = read();
if(x == 0) a[y] = z;
else b[y] = z;
} else {
l1 = read(), r1 = read(), l2 = read(), r2 = read();
cnt = 0;
while(l1 <= r1 || l2 <= r2) {
if(l1 <= r1 && l2 <= r2) {
if(a[l1] < b[l2]) now[++cnt] = a[l1++];
else now[++cnt] = b[l2++];
} else if(l1 <= r1) now[++cnt] = a[l1++];
else now[++cnt] = b[l2++];
}
// cout << "now: ";
// for(int i = 1; i <= cnt; i++) cout << now[i] << " ";
// cout << '\n';
ans[++tot] = now[cnt / 2 + 1];
}
}
for(int i = 1; i < tot; i++) cout << ans[i] << '\n';
cout << ans[tot];
}
int main() {
freopen("median.in", "r", stdin);
freopen("median.out", "w", stdout);
n = read(), m = read();
for(int i = 1; i <= n; i++) a[i] = read();
for(int i = 1; i <= n; i++) b[i] = read();
if(n <= 1000 && m <= 2000) return solve(), 0;
return 0;
}
跑五秒才能过的正解代码
/*
By:Loceaner
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
return x * f;
}
const int N = 5e5 + 11;
int n, m, a[N], b[N], now[N], cnt, tot, ans[N];
int opt, x, y, z, l1, l2, r1, r2;
void solve() {
while(m--) {
opt = read();
if(opt == 1) {
x = read(), y = read(), z = read();
if(x == 0) a[y] = z;
else b[y] = z;
} else {
l1 = read(), r1 = read(), l2 = read(), r2 = read();
cnt = 0;
while(l1 <= r1 || l2 <= r2) {
if(l1 <= r1 && l2 <= r2) {
if(a[l1] < b[l2]) now[++cnt] = a[l1++];
else now[++cnt] = b[l2++];
} else if(l1 <= r1) now[++cnt] = a[l1++];
else now[++cnt] = b[l2++];
}
// cout << "now: ";
// for(int i = 1; i <= cnt; i++) cout << now[i] << " ";
// cout << '\n';
ans[++tot] = now[cnt / 2 + 1];
}
}
for(int i = 1; i < tot; i++) cout << ans[i] << '\n';
cout << ans[tot];
}
int main() {
freopen("median.in", "r", stdin);
freopen("median.out", "w", stdout);
n = read(), m = read();
int maxn = 0;
for(int i = 1; i <= n; i++) a[i] = read(), maxn = max(maxn, a[i]);
for(int i = 1; i <= n; i++) b[i] = read(), maxn = max(maxn, a[i]);
if(n <= 1000 && m <= 2000) return solve(), 0;
while(m--) {
opt = read();
if(opt == 1) {
x = read(), y = read(), z = read();
if(!x) a[y] = z;
else b[y] = z;
maxn = max(maxn, z);
} else if(opt == 2) {
l1 = read(), r1 = read(), l2 = read(), r2 = read();
int l = 0, r = maxn, len = (r1 - l1 + 1 + r2 - l2 + 1) / 2 + 1, ans;
while(l <= r) {
int mid = (l + r) >> 1;
int now1 = upper_bound(a + l1, a + r1 + 1, mid) - (a + l1);
int now2 = upper_bound(b + l2, b + r2 + 1, mid) - (b + l2);
if(now1 + now2 >= len) {
ans = mid;
r = mid - 1;
r1 = l1 + now1 - 1;
r2 = l2 + now2 - 1;
}
else {
l = mid + 1;
len -= now1 + now2;
l1 = l1 + now1;
l2 = l2 + now2;
}
}
cout << ans << '\n';
}
}
return 0;
}正解\(1\)(要九九八，不要九十八，只需一点五，AC带回家)
#include<cstdio>
#include<algorithm>
const int N = 5e5 + 10;
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 a[N], b[N];
int Cal(int l, int r, int *a, int st, int ed, int *b) {
int p = r - l + 1 + ed - st + 1 >> 1, L = l, R = r;
for(;L <= R;) {
int m = L + R >> 1, c = p - (m - l + 1) + st;
if(c == st - 1 && m - l == p && a[m] <= b[st]) return a[m];
if(c < st) {R = m - 1; continue;}
else if(c > ed) {L = m + 1; continue;}
if(a[m] >= b[c] && (a[m] <= b[c + 1] || c == ed)) return a[m];
if(a[m] >= b[c]) R = m - 1;
else L = m + 1;
}
return 0;
}
int main() {
freopen("median.in","r",stdin);
freopen("median.out","w",stdout);
int n = ri(), m = ri();
for(int i = 1;i <= n; ++i) a[i] = ri();
for(int i = 1;i <= n; ++i) b[i] = ri();
for(;m--;) {
int op = ri();
if(op == 1) {
int w = ri(), x = ri(), y = ri();
!w ? a[x] = y : b[x] = y;
}
else {
int l = ri(), r = ri(), l1 = ri(), r1 = ri(), x;
if(x = Cal(l, r, a, l1, r1, b)) printf("%d\n", x);
else printf("%d\n", Cal(l1, r1, b, l, r, a));
}
}
return 0;
}
/*
6 1
1 2 4 5 6 7
1 1 3 5 6 7
2 1 6 1 1
*/正解\(2\)
#include <bits/stdc++.h>
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;
}
const int N = 5e5+50;
int a[N], b[N];
int kth(int ta[], int sa, int tb[], int sb, int k) {
if (sa > sb) return kth(tb, sb, ta, sa, k);
if (sa == 0) return tb[k];
if (k == 1) return std::min(ta[1], tb[1]);
int ka = std::min(sa, k/2), kb = k - ka;
if (ta[ka] < tb[kb]) return kth(ta+ka, sa-ka, tb, sb, k-ka);
return kth(ta, sa, tb+kb, sb-kb, k-kb);
}
int query(int la, int ra, int lb, int rb) {
int sa = ra-la+1, sb = rb-lb+1, siz = sa + sb;
return kth(a+la-1, sa, b+lb-1, sb, siz/2+1);
}
int main() {
freopen("median.in", "r", stdin);
freopen("median.out", "w", stdout);
int n, m;
n = ri(); m = ri();
for (int i = 1; i <= n; i++) a[i] = ri();
for (int i = 1; i <= n; i++) b[i] = ri();
for (int opt; m--; ) {
opt = ri();
if (opt == 2) {
int la = ri(), ra = ri(), lb = ri(), rb = ri();
printf("%d\n", query(la, ra, lb, rb));
} else {
int p = ri(), pos = ri(), val = ri();
if (p == 0) a[pos] = val;
else b[pos] = val;
}
}
return 0;
}
/*
5 5
12 41 46 68 69
35 61 82 84 96
2 1 4 3 5
1 0 5 75
2 2 4 3 4
2 3 4 1 5
2 1 4 2 4
*/
考场爆零代码
/*
By:Loceaner
*/
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#define ll long long
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
return x * f;
}
const int N = 2e5 + 11;
int n, A, B, C, D;
int a[N];
ll f[N], dp[N];
ll calc(int x) {
return 1LL * A * x * x * x + B * x * x + C * x + D;
}
int minn[1011][1011];
inline int query(int l, int r) {
int k = log2(r - l + 1);
return min(minn[l][k], minn[r - (1 << k) + 1][k]);
}
void solve1() {
memset(dp, -0x3f, sizeof(dp));
dp[0] = 0;
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= i; j++) {
dp[i] = max(dp[i], dp[j - 1] + calc(query(j, i)));
}
}
cout << dp[n] << '\n';
return;
}
void solve2() {
memset(dp, -0x3f, sizeof(dp));
dp[0] = 0;
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
dp[i] = max(dp[i], dp[j - 1] + calc(minn[j][i]));
}
}
cout << dp[n] << '\n';
return;
}
int main() {
freopen("min.in", "r", stdin);
freopen("min.out", "w", stdout);
n = read(), A = read(), B = read(), C = read(), D = read();
for(int i = 1; i <= n; i++) a[i] = read(), f[i] = calc(a[i]), minn[i][0] = a[i];
for(int j = 1; (1 << j) <= n; j++)
for(int i = 1; i <= n; i++)
minn[i][j] = min(minn[i][j - 1], minn[i + (1 << (j - 1))][j - 1]);
// for(int i = 1; i <= n; i++) cout << f[i] << ' ';
if(n <= 1000) return solve1(), 0;
if(A == 0 && B == 0 && C <= 0) return solve2(), 0;
else for(int i = 1;i <= n; ++i) if(fabs(calc(a[i])) >= 1e12) {return printf("%d\n", a[i]), 0;};
return 0;
}
正解
#include<cstdio>
#include<cstring>
#include<algorithm>
const int Nt = 524287; const long long inf = 0x3f3f3f3f3f3f3f3f;
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 a[Nt], st[Nt], tp, n, A, B, C, D;
long long T[Nt << 1], f[Nt], mx[Nt];
void Up(int i, long long x)
{for(T[i += Nt] = x; i >>= 1;) T[i] = std::max(T[i << 1], T[i << 1 | 1]);}
long long Cal(long long x) {return ((A * x + B) * x + C) * x + D;}
int main() {
freopen("min.in","r",stdin);
freopen("min.out","w",stdout);
n = ri(); A = ri(); B = ri(); C = ri(); D = ri();
for(int i = 1;i <= n; ++i) a[i] = ri();
std::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]; long long x = f[i];
for(;st[tp] > a[i + 1] && tp;) x = std::max(x, mx[tp]), Up(tp--, -inf);
st[++tp] = a[i + 1]; mx[tp] = x; Up(tp, x + Cal(st[tp]));
}
printf("%lld\n", f[n]);
return 0;
}神仙正解
#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;
}标签：最优 解题报告 端点 isp val mat read main efi
原文地址：https://www.cnblogs.com/loceaner/p/11679310.html
