解题报告 『CF896C Willem, Chtholly and Seniorious（ODT）』

原题地址

国内珂朵莉树的起源，同时也是板子题。

网上就能找到很多资料，在此我就不赘述了。

 

代码实现如下：

 

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define IT set<node>::iterator
#define rep(i, a, b) for (register int i = (a); i <= (b); i++)

const int mod7 = 1e9 + 7, maxn = 1e5 + 5;

int n, m, seed, vmax;
int a[maxn];

vector< pair<int, int> > v;

struct node {
    int l, r;
    mutable int v;
    node(int L, int R = -1, int V = 0): l(L), r(R), v(V) {}
    int operator <(const node &o) const {
        return l < o.l;
    }
};

set<node> s;

int read() {
    int x = 0, flag = 0;
    char ch = ' ';
    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if (ch == '-') {
        flag = 1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + (ch ^ '0');
        ch = getchar();
    }
    return flag ? -x : x;
}

int rnd() {
    int ret = seed;
    seed = (seed * 7 + 13) % mod7;
    return ret;
}

int quick_power(int a, int b, int mod)  {
    int ans = 1, base = a % mod;
    while (b > 0) {
        if (b & 1) {
            ans *= base;
            ans %= mod;
        }
        b >>= 1;
        base *= base;
        base %= mod;
    }
    return ans %= mod;
}

IT split(int pos) {
    IT it = s.lower_bound(node(pos));
    if (it != s.end() && it->l == pos) return it;
    it--;
    int L = it->l, R = it->r, V = it->v;
    s.erase(it);
    s.insert(node(L, pos - 1, V));
    return s.insert(node(pos, R, V)).first;
}

void add(int l, int r, int val) {
    IT itr = split(r + 1), itl = split(l);
    while (itl != itr) {
        itl->v += val;
        itl++;
    }
}

void assign(int l, int r, int val) {
    IT itr = split(r + 1), itl = split(l);
    s.erase(itl, itr);
    s.insert(node(l, r, val));
}

int kth(int l, int r, int k) {
    IT itr = split(r + 1), itl = split(l);
    v.clear();
    while (itl != itr) {
        v.push_back(pair<int, int>(itl->v, itl->r - itl->l + 1));
        itl++;
    }
    sort(v.begin(), v.end());
    for (vector< pair<int, int> >::iterator it = v.begin(); it != v.end (); it++) {
        k -= it->second;
        if (k <= 0) return it->first;
    }
    return -1;
}

int sum(int l, int r, int ex, int mod) {
    IT itr = split(r + 1), itl = split(l);
    int res = 0;
    while (itl != itr) {
        res = (res + (itl->r - itl->l + 1) * quick_power(itl->v, ex, mod)) % mod;
        itl++;
    }
    return res;
}

void write(int x) {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}

signed main() {
    n = read(), m = read(), seed = read(), vmax = read();
    rep(i, 1, n) {
        a[i] = (rnd() % vmax) + 1;
        s.insert(node(i, i, a[i]));
    }
    s.insert(node(n + 1, n + 1, 0));
    rep(i, 1, m) {
        int opt = (rnd() % 4) + 1;
        int l = (rnd() % n) + 1;
        int r = (rnd() % n) + 1;
        if (l > r) l ^= r ^= l ^= r;
        int x, y;
        if (opt == 3) x = (rnd() % (r - l + 1)) + 1;
        else x = (rnd() % vmax) + 1;
        if (opt == 4) y = (rnd() % vmax) + 1;
        if (opt == 1) add(l, r, x);
        else if (opt == 2) assign(l, r, x);
        else if (opt == 3) {
            write(kth(l, r, x));
            printf("\n");
        }
        else {
            write(sum(l, r, x, y));
            printf("\n");
        }
    }
    return 0;
}