ZYB's Premutation POJ5592

Problem Description

ZYBZYBZYB
has a premutation PPP,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)(i,j)(i < j)(i,j)(i<j)
is considered as a reverse log if Ai>AjA_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 NNN.

In the next line there are NNN
numbers AiA_iA​i​​,describe
the number of the reverse logs of each prefix,

The input is correct.

1≤T≤51 \leq T \leq 51≤T≤5,1≤N≤500001
\leq N \leq 500001≤N≤50000

Output

For each testcase,print the ans.

Sample Input

1
3
0 1 2

Sample Output

3 1 2

问题描述

ZYBZYBZYB有一个排列PPP,但他只记得PPP中每个前缀区间的逆序对数,现在他要求你还原这个排列.

(i,j)(i<j)(i,j)(i < j)(i,j)(i<j)被称为一对逆序对当且仅当Ai>AjA_i>A_jA​i​​>A​j​​

输入描述

第一行一个整数TTT表示数据组数。

接下来每组数据：

第一行一个正整数NNN,描述排列的长度.

第二行NNN个正整数AiA_iA​i​​,描述前缀区间[1,i][1,i][1,i]的逆序对数.

数据保证合法.

1≤T≤51 \leq T \leq 51≤T≤5,1≤N≤500001 \leq N \leq 500001≤N≤50000

输出描述

TTT行每行NNN个整数表示答案的排列.

输入样例

1
3
0 1 2

输出样例

3 1 2

超时：：：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
int main ()
{
    int T,n;
    scanf("%d",&T);
    long long int arr[50010];
    int mark[50010],cnt;
    int res[50010];
    bool vis[50010];
    for(int xx=0; xx<T; xx++)
    {
        memset(vis,0,sizeof(vis));
        cnt = 0;
        int n;
        long long temp;
        scanf("%d",&n);
        if(n>=1)
           scanf("%I64d",arr);
        long long int xxx=0;
        for(int i=1; i<n; i++){
            scanf("%I64d",&temp);
            arr[i] = temp-xxx,xxx=temp;
        }
        for(int i=0; i<n; i++)
        {
            for(int i=n-1; i>=0; i--)
            {
                if(arr[i]==0&&!vis[i])
                {
                    mark[cnt++] = i;
                    arr[i]--;
                    vis[i]=1;
                    break;
                }
                else arr[i]--;
            }
        }
        int mmax = n;
        for(int i=0; i<n; i++,mmax--)
            res[mark[i]] = mmax;
        for(int i=0; i<n; i++){
            if(i!=n-1) printf("%d ",res[i]);
             else printf("%d",res[i]);
        }
        printf("\n");
    }
    return 0;
}