「2019纪中集训Day18」解题报告

T1、完全背包

\(n \ (n \leq 10 ^ 6)\) 件物品,体积为 \(a_i \ (a_i \leq 100)\),价值为 \(b_i \ (b_i \leq 100)\)。求一个容量为 \(m \ (m \leq 10 ^ {18})\) 的背包可获得的最大价值。

\(Sol_1\)

\(lemma\):任意 \(n\) 个整数中一定能取出一段数使得它们的和被 \(n\) 整除。
\(prf\)
\(n\) 个数的前缀和为 \(S_i\)
\(\{i \in [0,n] \bigcap \Z \ | \ S_i \}\) 里的 \(n + 1\) 个数,一定存在一组 \(i,j\) 满足 \(S_i \equiv S_j \ mod \ n\)
\(Q.E.D.\)

观察到物品只需保留不超过 \(100\) 个。

设所有物品中性价比最高的物品编号为 \(s\)
则最优方案中,性价比小于 \(\frac{b_s}{a_s}\) 的物品数量不超过 \(a_s\)
证明可以用到上述引理,若个数超过 \(a_s\) 一定可以选出来一些用物品 \(s\) 替换。
所以超过 \(100a_s\) 的部分全部用来取第 \(s\) 件物品,剩下的完全背包即可。

复杂度 \(O(100 \times 100 a_s)\)

\(Source_1\)

#include <cstdio>
#include <cstring>
#include <algorithm>
int in() {
    int x = 0; char c = getchar(); bool f = 0;
    while (c < '0' || c > '9')
        f |= c == '-', c = getchar();
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? -x : x;
}
long long lin() {
    long long x = 0; char c = getchar(); bool f = 0;
    while (c < '0' || c > '9')
        f |= c == '-', c = getchar();
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? -x : x;
}
template<typename T>inline void chk_min(T &_, T __) { _ = _ < __ ? _ : __; }
template<typename T>inline void chk_max(T &_, T __) { _ = _ > __ ? _ : __; }

const int N = 1e4;

struct node {
    int a, b;
    inline bool operator < (const node& y) const {
        if (this->b * y.a == y.b * this->a)
            return this->a < this->a;
        return this->b * y.a > y.b * this->a;
    }
} t[105];
int max[105];

int n, nn, f[N + N + 5];
long long m, res;

void backpack() {
    f[0] = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = t[i].a; j <= nn; ++j)
            chk_max(f[j], f[j - t[i].a] + t[i].b);
}

int main() {
    //freopen("in", "r", stdin);
    freopen("backpack.in", "r", stdin);
    freopen("backpack.out", "w", stdout);
    n = in(), m = lin();
    memset(max, -1, sizeof(max));
    for (int i = 1, x, y; i <= n; ++i) {
        x = in(), y = in();
        chk_max(max[x], y);
    }
    n = 0;
    for (int i = 1; i <= 100; ++i)
        if (~max[i])
            t[++n] = (node){i, max[i]};
    std::sort(t + 1, t + 1 + n);

    if (m > N) {
        res = (m - N) / t[1].a * t[1].b;
        nn = (m - N) % t[1].a + N;
    } else {
        nn = m;
    }
    backpack();
    res += f[nn];

    printf("%lld\n", res);
    return 0;
}

\(Sol_2\)

这种数据范围很容易想到用矩阵来优化,可以试着构造一种矩阵运算 \(*\)
设有两个矩阵 \(A, B\),且 \(A * B = C\)
我们规定 \(C_{i, j} = \max_{k = 1} ^ {n} \{ A_{i, k} + B_{k, j} \}\) (类似矩阵乘法)。
该运算的单位矩阵为:
\[ \left[ \begin{matrix} 0 & -\infin & -\infin & \cdots & -\infin & -\infin \\ -\infin & 0 & -\infin & \cdots & -\infin & -\infin \\ -\infin & -\infin & 0 & \cdots & -\infin & -\infin \\ \vdots & \vdots & \vdots & \ddots & 0 & -\infin \\ -\infin & -\infin & -\infin & \cdots & -\infin & 0 \end{matrix} \right] \]
并不难证明该运算有结合律,故快速 \(*\) (???)

时间复杂度 \(O(100 ^ 3 \log_2 m)\)

\(Source_2\)

//#pragma GCC optimize(3,"Ofast","inline")
#include <cstdio>
#include <cstring>
#include <algorithm>
int in() {
    int x = 0; char c = getchar(); bool f = 0;
    while (c < '0' || c > '9')
        f |= c == '-', c = getchar();
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? -x : x;
}
long long lin() {
    long long x = 0; char c = getchar(); bool f = 0;
    while (c < '0' || c > '9')
        f |= c == '-', c = getchar();
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? -x : x;
}
template<typename T>inline void chk_min(T &_, T __) { _ = _ < __ ? _ : __; }
template<typename T>inline void chk_max(T &_, T __) { _ = _ > __ ? _ : __; }

const int N = 105;

int max[105], n;
long long m;

struct matrix {
    long long a[N][N];
    matrix(long long _ = -1) {
        memset(a, -1, sizeof(a));
        if (~_)
            for (int i = 0; i < 100; ++i)
                a[i][i] = _;
    }
    inline long long* operator [] (const int x) {
        return a[x];
    }
    matrix operator * (matrix b) const {
        matrix ret;
        for (int k = 0; k < 100; ++k)
            for (int i = 0; i < 100; ++i)
                if (~a[i][k])
                    for (int j = 0; j < 100; ++j)
                        if (~b[k][j])
                            chk_max(ret[i][j], a[i][k] + b[k][j]);
        return ret;
    }
    matrix operator ^ (long long b) const {
        matrix ret(0), base = *this;
        for (; b; b >>= 1ll, base = base * base)
            if (b & 1ll)
                ret = ret * base;
        return ret;
    }
} ;

void backpack(long long *f) {
    f[0] = 0;
    for (int i = 1; i <= 100; ++i) {
        if (~max[i]) {
            for (int j = i; j < 100; ++j) {
                if (~f[j - i]) {
                    chk_max(f[j], f[j - i] + max[i]);
                }
            }
        }
    }
}

int main() {
    //freopen("in", "r", stdin);
    freopen("backpack.in", "r", stdin);
    freopen("backpack.out", "w", stdout);
    n = in(), m = lin();
    memset(max, -1, sizeof(max));
    for (int i = 1, a, b; i <= n; ++i) {
        a = in(), b = in();
        chk_max(max[a], b); 
    }
    matrix f, trans;
    backpack(f[0]);
    for (int i = 0; i < 99; ++i)
        trans[i + 1][i] = 0;
    for (int i = 0; i < 100; ++i)
        trans[i][99] = max[100 - i];
    f = f * (trans ^ m);
    printf("%lld\n", f[0][0]);
    return 0;
}

T2、中间值

有两个长度为 \(n \ (n \leq 5 \times 10 ^ 5)\) 的非降序列 \(a,b\)\(m \ (m \leq 10 ^ 6)\) 次操作:
1、修改 \(a(0)\)\(b(1)\) 中的某个数,保证修改后依旧非降;
2、求将某两个区间 \([l_a,r_a], [l_b,r_b]\) 合并后中间的数,保证两区间长度和为奇数。

\(Sol_1\)

直接二分,细节较多;
比如在 \(l_a, r_a\) 中二分到 \(mid\),只需判断 \(a_{mid}\)\(b\) 中应该出现的位置是否合法即可。

\(Sol_2\)

考虑分治。
设当前要求第 \(k\) 大元素,可能出现答案的区间为 \([l_a, r_a], [l_b, r_b]\)
设两个区间中第 \(\frac{k}{2}\) 大的元素为 \(a_x, b_y\) (不讨论边界);
不妨设 \(a_x < b_y\),那么 \(a\) 中在 \(x\) 之前、\(b\) 中在 \(y\) 之后的数就不是答案,递归即可。

以上两种做法都可以扩展到第 \(k\) 大,做法同上。
时间复杂度都为 \(O(m \log_2 n)\)

\(Source\)

//#pragma GCC optimize(3,"Ofast","inline")
#include <cstdio>
#include <algorithm>
int in() {
    int x = 0; char c = getchar(); bool f = 0;
    while (c < '0' || c > '9')
        f |= c == '-', c = getchar();
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? -x : x;
}
template<typename T>inline void chk_min(T &_, T __) { _ = _ < __ ? _ : __; }
template<typename T>inline void chk_max(T &_, T __) { _ = _ > __ ? _ : __; }

const int N = 5e5 + 5;

int n, m, a[N], b[N];

int binary_search(int xl, int xr, int yl, int yr, int *x, int *y) {
    int l = xl, r = xr, mid, t, k = (xr - xl + yr - yl + 3) >> 1;
    while (l < r) {
        mid = (l + r) >> 1;
        t = (k - 1) - (mid - xl) + yl;
        if (t < yl) {
            r = mid - 1;
            continue;
        } if (t > yr + 1) {
            l = mid + 1;
            continue;
        }
        if ((t == yr + 1 || y[t] >= x[mid]) && (t == yl || y[t - 1] <= x[mid]))
            return mid;
        if (y[t] < x[mid]) {
            r = mid - 1;
        } else if (y[t - 1] > x[mid]) {
            l = mid + 1;
        }
    }
    if (l == r) {
        t = (k - 1) - (l - xl) + yl;
        if ((t == yr + 1 || y[t] >= x[l]) && (t == yl || y[t - 1] <= x[l]))
            return l;
    }
    return -1;
}

int calc(int l1, int r1, int l2, int r2, int k) {
    if (l1 > r1)
        return b[l2 + k - 1];
    if (l2 > r2)
        return a[l1 + k - 1];
    if (k == 1)
        return a[l1] < b[l2] ? a[l1] : b[l2];
    int p1, p2;
    if (r1 - l1 + 1 < (k >> 1)) {
        p1 = r1 - l1 + 1;
        p2 = k - p1;
    } else if (r2 - l2 + 1 < ((k + 1) >> 1)) {
        p2 = r2 - l2 + 1;
        p1 = k - p2;
    } else {
        p1 = k >> 1;
        p2 = (k + 1) >> 1;
    }
    if (a[l1 + p1 - 1] < b[l2 + p2 - 1]) {
        return calc(l1 + p1, r1, l2, l2 + p2 - 1, p2);
    } else {
        return calc(l1, l1 + p1 - 1, l2 + p2, r2, p1);
    }
}

int main() {
    //freopen("in", "r", stdin);
    //freopen("out", "w", stdout);
    freopen("median.in", "r", stdin);
    freopen("median.out", "w", stdout);
    n = in(), m = in();
    for (int i = 1; i <= n; ++i)
        a[i] = in();
    for (int i = 1; i <= n; ++i)
        b[i] = in();

    int la, ra, lb, rb, res;
    while (m--) {
        if (in() == 1) {
            if (!in()) {
                la = in();
                a[la] = in();
            } else {
                la = in();
                b[la] = in();
            }
        } else {
            la = in(), ra = in(), lb = in(), rb = in();
            //printf("%d\n", calc(la, ra, lb, rb, (ra - la + rb - lb + 3) >> 1));
            res = binary_search(la, ra, lb, rb, a, b);
            if (~res) {
                printf("%d\n", a[res]);
            } else {
                res = binary_search(lb, rb, la, ra, b, a);
                if (~res) {
                    printf("%d\n", b[res]);
                }
            }
        }
    }
    return 0;
}

T3、Sequence

\(f(A)\) 表示所有长度为 \(n \ (n\leq 10 ^ {18})\) 且满足 \(\forall i \in [1,n] \ a_i | A\) 的序列的价值和;
一个序列的价值为 \(gcd(a_1, a_2, a_3 \ \dots \ a_n, B), \ (B \leq 10 ^ {18})\)
\(\sum_{i = 1}^m f(i), \ (m \leq 2 \times 10 ^ {7})\),答案对 \(998244353\) 取模。

\(Sol\)

显然地,若 \(gcd(x, y) = 1\),则有 \(gcd(xy, B) = gcd(x, B) \times gcd(y, B)\)
同样地,若 \(gcd(x, y) = 1\),则 \(f(xy)\) 所对应的集合为 \(f(x)\)\(f(y)\) 对应集合的笛卡尔积。
\(f(x)\) 是一个积性函数。

只需计算形如 \(x = p ^ c\)\(f(x)\) 即可,其中 \(p\) 是质数;
\(k\) 满足 \(p ^ k | B\)\(p ^ {k + 1} \nmid B\),那么有:
\[ \begin{align} f(x) &= \sum_{i = 0}^{c} p ^ i \big[ (c - i + 1) ^ n - (c - i) ^ n \big] , \ (c \leq k) \notag \\ f(x) &= p ^ k (c - k + 1) ^ n + \sum_{i = 0}^{k - 1} p ^ i \big[ (c - i + 1) ^ n - (c - i) ^ n \big] , \ (c > k) \notag \\ \end{align} \]
时间复杂度 \(O(\frac{m}{\ln m} \times\log_? ^2 m)\),计算 \(f(p ^ c)\) 时需要枚举 \(c\),这个 \(\log\) 很小可以当做常数对待 (应该)

\(Source\)

//#pragma GCC optimize(3,"Ofast","inline")
#include <cstdio>
#include <algorithm>
int in() {
    int x = 0; char c = getchar(); bool f = 0;
    while (c < '0' || c > '9')
        f |= c == '-', c = getchar();
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? -x : x;
}
long long lin() {
    long long x = 0; char c = getchar(); bool f = 0;
    while (c < '0' || c > '9')
        f |= c == '-', c = getchar();
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? -x : x;
}
template<typename T>inline void chk_min(T &_, T __) { _ = _ < __ ? _ : __; }
template<typename T>inline void chk_max(T &_, T __) { _ = _ > __ ? _ : __; }

const int N = 2e7 + 5, mod = 998244353;

long long n, B;
int m, pri_cnt, pri[N / 10], min_pri[N], f[N], Pow[105];
std::pair<int, int> g[N];

inline void add(int &_, int __) {
    _ += __;
    if (_ >= mod)
        _ -= mod;
}

int qpow(int base, int b, int ret = 1) {
    for (; b; b >>= 1, base = 1ll * base * base % mod)
        if (b & 1)
            ret = 1ll * ret * base % mod;
    return ret;
}

void Euler_sieve() {
    for (int i = 2; i <= m; ++i) {
        if (!min_pri[i]) {
            g[i].second = min_pri[i] = pri[++pri_cnt] = i;
            g[i].first = 1;
        }
        for (int j = 1, tmp; j <= pri_cnt && i * pri[j] <= m; ++j) {
            tmp = i * pri[j];
            g[tmp].second = min_pri[tmp] = pri[j];
            g[tmp].first = 1;
            if (!(i % pri[j])) {
                g[tmp].first += g[i].first;
                g[tmp].second *= g[i].second;
                break;
            }
        }
    }
}

void calc_f() {
    int nn = n % (mod - 1);
    for (int i = 0; i <= 100; ++i)
        Pow[i] = qpow(i, nn);
    f[1] = 1;
    for (int i = 2; i <= m; ++i) {
        if (g[i].second == i) {
            for (int j = 0, tmp = 1; j <= g[i].first; ++j) {
                add(f[i], 1ll * tmp * (Pow[g[i].first - j + 1] - Pow[g[i].first - j] + mod) % mod);
                if (!((B / tmp) % min_pri[i]))
                    tmp *= min_pri[i];
            }
        } else {
            f[i] = 1ll * f[i / g[i].second] * f[g[i].second] % mod;
        }
    }
    for (int i = 1; i <= m; ++i)
        add(f[i], f[i - 1]);
}

int main() {
    //freopen("in", "r", stdin);
    freopen("sequence.in", "r", stdin);
    freopen("sequence.out", "w", stdout);
    n = lin(), m = in(), B = lin();
    Euler_sieve();
    calc_f();
    printf("%d\n", f[m]);
    return 0;
}
posted @ 2019-08-19 20:52  15owzLy1  阅读(...)  评论(... 编辑 收藏