题意:一共有$n$个数,$m$次询问,每次询问包括$k、pos$两个数,需要你从这$n$个数里面找出$k$个数,使得他们的总和最大,如果有多种情况,找出序号字典序最小的一组,然后输出这个序列中第$pos$个数的值。

思路:根据贪心的思想,把这$n$个数按大到小排个序$($相同大小按下标从小到大$)$,再按照每个元素的下标依次放入主席树中,对于每次询问,查找前$k$个数中第$pos$大的数,但这个查找到的值是下标,再把这个下标对应到数值即可,体现在代码中的$b[]$数组中。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector> using namespace std; const int N = ; struct date {
int val, pos;
}; struct node {
int l, r, sum;
}; int n, q, cnt, root[N], b[N];
date a[N];
node tree[ * N]; bool cmp(date one, date two)
{
if (one.val != two.val) return one.val > two.val;
return one.pos < two.pos;
} void update(int l, int r, int &x, int y, int pos)
{
tree[++cnt] = tree[y];
tree[cnt].sum++, x = cnt;
if (l == r) return;
int mid = (l + r) / ;
if (mid >= pos) update(l, mid, tree[x].l, tree[y].l, pos);
else update(mid + , r, tree[x].r, tree[y].r, pos);
} int query(int l, int r, int x, int y, int k)
{
if (l == r) return l;
int mid = (l + r) / ;
int sum = tree[tree[y].l].sum - tree[tree[x].l].sum;
if (sum >= k) return query(l, mid, tree[x].l, tree[y].l, k);
else return query(mid + , r, tree[x].r, tree[y].r, k - sum);
} int main()
{
scanf("%d", &n);
for (int i = ; i <= n; i++) {
scanf("%d", &a[i].val);
a[i].pos = i;
}
sort(a + , a + n + , cmp);
for (int i = ; i <= n; i++) {
b[a[i].pos] = a[i].val;
update(, n, root[i], root[i - ], a[i].pos);
}
scanf("%d", &q);
while (q--) {
int k, pos;
scanf("%d%d", &k, &pos);
printf("%d\n", b[query(, n, root[], root[k], pos)]);
}
return ;
}

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