ZYB's Premutation POJ5592

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 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 $N$.

In the next line there are $N$ numbers AiA_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

1
3
0 1 2

Sample Output

3 1 2

问题描述

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

$(i,j)(i<j)$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$T$表示数据组数。

接下来每组数据：

第一行一个正整数$N$,描述排列的长度.

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

数据保证合法.

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

输出描述

$T$行每行$N$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("
");
    }
    return 0;
}