2019.10.15考试解题报告

Posted 2021-03-26 loceaner

总结

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

---

思路

T1

先打(30)分暴力，这个很好打

然后(70)分(O(nlog^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 (zle 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

T2

(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。

T3

不会，咕咕咕

---

代码

T1

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 << '
';
            ans[++tot] = now[cnt / 2 + 1];
        }
    }
    for(int i = 1; i < tot; i++) cout << ans[i] << '
';
    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 << '
';
            ans[++tot] = now[cnt / 2 + 1];
        }
    }
    for(int i = 1; i < tot; i++) cout << ans[i] << '
';
    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 << '
';
        }
    }
    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
", x);
            else printf("%d
", 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
", 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
*/

T2

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] << '
';
    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] << '
';
    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
", 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
", f[n]);
    return 0;
}

T3

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
", (A + P) % P);
    return 0;
}