联赛模拟测试32

T2

20Pts

• \(n^2\)，暴力即可，注意卡常，max可以define一下

#include <bits/stdc++.h>
using namespace std;
#define max(a, b) (a) > (b) ? (a) : (b)
inline int read() {
    int k = 0, f = 1; char ch = getchar();
    for (; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
    for (; isdigit(ch); ch = getchar()) k = k * 10 + ch - '0';
    return k * f;
}
const int maxn = 5e5 + 100;
struct node { int a, b; } a[maxn];
int main() {
    freopen("A.in", "r", stdin);
    freopen("A.out", "w", stdout);
    register int n = read(), q = read();
    for (register int i = 1; i <= n; i++) a[i].a = read(), a[i].b = read();
    while (q--) {
        register int x = read(); register long long ans = -0x3f3f3f3f3f3f3f;
        for (register int i = 1; i <= n; i++) { long long tmp = 1ll * (1ll * x * x * a[i].a + 1ll * x * a[i].b); ans = max(ans, tmp); }
        printf("%lld\n", ans);
    }
}

40pts

• 按照数据范围\(x\)的\(|32323|\)显然比\(q\)的\(5e5\)小好多，开一个O2，题库能过，估计不开的话评测机也是可以的。

#include <bits/stdc++.h>
using namespace std;
#define max(a, b) (a) > (b) ? (a) : (b)
inline int read() {
    int k = 0, f = 1; char ch = getchar();
    for (; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
    for (; isdigit(ch); ch = getchar()) k = k * 10 + ch - '0';
    return k * f;
}
const int maxn = 5e5 + 100;
struct node { int a, b; } a[maxn];
long long ans[32324];
long long absans[32324];
int main() {
    freopen("A.in", "r", stdin);
    freopen("A.out", "w", stdout);
    register int n = read(), q = read();
    register int x;
    for (register int i = 1; i <= n; i++) a[i].a = read(), a[i].b = read();
    register long long cur1, cur2, tmp1, tmp2;
    for (register int j = 0; j <= 32323; j++) {
        cur1 = cur2 = -0x7f7f7f7f;
        for (register int i = 1; i <= n; i++) {
            tmp1 = 1ll * j * j * a[i].a + 1ll * j * a[i].b;
            tmp2 = 1ll * j * j * a[i].a - 1ll * j * a[i].b;
            cur1 = max(cur1, tmp1), cur2 = max(cur2, tmp2);
        }
        ans[j] = cur1;
        absans[j] = cur2;
    }
    while (q--) {
        x = read();
        if (x > 0) printf("%lld\n", ans[x]);
        else printf ("%lld\n", absans[-x]);
    }
}

T3:

• 首先第一个数据点, \(n <= 5\)，记忆化搜索。
• 第二个数据点，找规律：

    int jc = 1, jc2 = 1;
    int tmp = 0;
    for (int i = n; i >= 1; i--) {
        jc = 1ll * jc * i % mod;
        if (i != n) jc2 = 1ll * jc2 * i % mod;
        tmp = (tmp + 1ll * qpow(jc, mod - 2) * jc2 % mod * 1ll * (n - i + 1) % mod) % mod;
    }
    cout << tmp % mod << endl;
    return 0;

• \(n == 2\)的时候，可以分为两种情况，第一种是不选完，这样的时候每次只可能

    int x = read(), y = read();
    jc[0] = ny[0] = ny[1] = jcny[0] = jcny[1] = 1;
    for (int i = 1; i <= x + y; i++) jc[i] = jc[i - 1] * i % mod;
    for (int i = 2; i <= x + y; i++) ny[i] = (mod - mod / i) * ny[mod % i] % mod;
    for (int i = 2; i <= x + y; i++) jcny[i] = jcny[i - 1] * ny[i] % mod;
    for (int i = x; i < x + y; i++) {
        int a = qpow(2, i);
        int b = qpow(a, mod - 2);
        ans += C(i - 1, x - 1) * b % mod * i % mod;
        //cout << C(i - 1, x - 1) << " " << a << endl;
    }
    for (int i = 0; i < x; i++) {
        int a = qpow(2, i + y);
        int b = qpow(a, mod - 2);
        ans += C(i + y - 1, i) * b % mod * (x + y) % mod;
        //cout << C(i + y - 1, i) << " " << a << endl;
    }
    cout << ans % mod << endl;

T4

• 直接暴力跳每一个点可过：1\2\8\9

    for (int i = 1; i <= q; i++) {
        int u = Q[i].u, v = Q[i].v;
        int LCA = lca(u, v), d = 0;
        long long ans = 0;
        int dis =  depth[u] + depth[v] - 2 * depth[LCA];
        for (; u != fa[LCA]; u = fa[u], d++) ans += (d | a[u]);
        d = 0;
        for (; v != LCA; v = fa[v], d++) ans += ((dis - d) | a[v]);
        printf("%lld\n", ans);
    }

• 当 \(a_i < 2\)， 的时候，因为只有偶数和 0 或的时候才会产生不一样的答案，所以使用倍增维护从每一个点 u 出发，与 u 距离为偶数并且点权为 1 的点的个数。