HDU 5592 ZYB's Premutation

Problem Description

$ZYB$ has a premutation $P$,but he only remeber the reverse log of each prefix of the premutation,now he ask you to restore the premutation.

Pair $(i,j)(i<j)$ is considered as a reverse log if A_i>A_jA​i​​>A​j​​ is matched.

Input

In the first line there is the number of testcases T.

For each teatcase:

In the first line there is one number $N$.

In the next line there are $N$ numbers A_iA​i​​,describe the number of the reverse logs of each prefix,

The input is correct.

$1\le T\le 5$,$1\le N\le 50000$

Output

For each testcase,print the ans.

Sample Input

130 1 2

Sample Output

3 1 2

根据逆序数求原序列，线段树优化一下

#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int maxn = 1e5 + 5;int T, n, m, a[maxn], b[maxn], ans[maxn];struct ST{    int f[maxn * 5];    void build(int x,int l,int r)    {        if (l == r) f[x] = 1;        else        {            int mid = l + r >> 1;            build(x + x, l, mid);            build(x + x + 1, mid + 1, r);            f[x] = f[x + x] + f[x + x + 1];        }    }    int find(int x, int l, int r, int v)    {        if (l == r) { f[x]--; return l; }        else        {            int mid = l + r >> 1;            f[x]--;            if (v > f[x + x + 1]) return find(x + x, l, mid, v - f[x + x + 1]);            else return find(x + x + 1, mid + 1, r, v);        }    }}st;int main(){    scanf("%d", &T);    while (T--)    {        scanf("%d", &n);        a[0] = 0;        st.build(1, 1, n);        for (int i = 1; i <= n; i++)        {            scanf("%d", &a[i]);            b[i] = a[i] - a[i - 1];        }        for (int i = n; i; i--)        {            ans[i] = st.find(1, 1, n, b[i] + 1);        }        for (int i = 1; i <= n; i++) printf("%d%s", ans[i], i == n ? "\n" : " ");    }    return 0;}