H、subseq ——树状数组

Kanade has an array a[1..n] , she define that an array b[1..m] is good if and only if it satisfy the following conditions: $1<=b[i]<=n b[i]<b[i+1]$for every i between 1 and m-1 a[b[i]] < a[b[i+1]] for every i between 1 and m-1 m>0 Now you need to find the k-th smallest lexicographically good array.

输入描述: The first line has two integer n,k

The second line has n integer a[i]

输出描述: If there is no solution, just only output -1, else output two lines, the first line has an integer m, the second line has m integer b[i]

备注: 1<=n <= 10^5

1<=k<=10^(18)

1<=a[i]<=10^9

示例 1 输入 3 2 1 2 3 输出 2 1 2

示例 2 输入 3 1000000000 1 2 3 输出 -1
因为这个是小于，所以查找的时候需要加一，然后在add的时候如果比k大要记成k+1，否则最后一个样例会超过ll的范围。。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=2000000;
ll a[maxn+5],f[maxn+5],c[maxn+5],tmp[maxn+5],ans[maxn+5];
int cnt;
ll n,k;
ll lowbit(ll x)
{
    return x&(-x);
}
void add(ll x,ll val)
{
    for(ll i=x;i>=1;i-=lowbit(i))
    {
        c[i]+=val;
        if(c[i]>k)
            c[i]=k+1;

    }
}
ll query(ll x)
{
    ll ans=0;
    for(ll i=x;i<=cnt;i+=lowbit(i))
    {
        ans+=c[i];
        if(ans>k)
            return k+1;
    }
    return ans;
}
int main()
{
    scanf("%lld%lld",&n,&k);
    for(ll i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        tmp[i]=a[i];
    }
    sort(tmp+1,tmp+1+n);
    cnt=unique(tmp+1,tmp+1+n)-(tmp+1);
    for(int i=1;i<=n;i++)
        a[i]=lower_bound(tmp+1,tmp+1+cnt,a[i])-tmp;
    for(int i=n;i>=1;i--)
    {
        f[i]=query(a[i]+1)+1;
        if(f[i]>k)
            f[i]=k+1;
        add(a[i],f[i]);
    }
    int ctans=0,maxn=0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]>maxn)
        {
            if(f[i]>=k)
                ans[++ctans]=i,k--,maxn=a[i];
            else
                k-=f[i];
        }
        if(k<=0)
            break;
    }
    if(k>0)
    {
        printf("-1\n");
        return 0;
    }
    printf("%d\n",ctans);
    for(int i=1;i<=ctans;i++)
        printf("%d%c",ans[i],i==ctans?'\n':' ');
    return 0;
}