P3988 [SHOI2013] 发牌

Solution

容易发现，答案就是维护当前序列的第 k 大值，而且只有删除，这个时候就可以使用权值线段树来维护。这颗树的每一个叶子表示一张牌，然后线段树记录改节点为根的子树的节点个数，接着进行查询即可，代码见下：

Code

#include <bits/stdc++.h>
#define ls(x) x << 1
#define rs(x) x << 1 | 1
#define IOS ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
using namespace std;
typedef long long ll;
struct SGT
{
    vector<int> tree;
    SGT(){}
    SGT(int n)
    {
        tree.resize(n << 2);
        build(1, 1, n);
    }
    void build(int u, int l, int r)
    {
        if(l == r)
        {
            tree[u] = 1;
            return;
        }
        int mid = l + r >> 1;
        build(ls(u), l, mid), build(rs(u), mid + 1, r);
        pushup(u);
        return;
    }
    void pushup(int u)
    {
        tree[u] = tree[ls(u)] + tree[rs(u)];
        return;
    }
    int query(int q, int u, int l, int r)
    {
        tree[u] --;
        if(l == r) return l;
        int mid = l + r >> 1;
        if(q <= tree[ls(u)]) return query(q, ls(u), l, mid);
        else return query(q - tree[ls(u)], rs(u), mid + 1, r);
    }
};
int n, tmp, now;
int main() 
{
    IOS;
    cin >> n;
    SGT sgtree(n);
    for(int i = 1; i <= n; i ++)
    {
        cin >> tmp;
        now = (now + tmp) % (n - i + 1);
        cout << sgtree.query(now + 1, 1, 1, n) << "\n";
    }
    return 0;
}