reverse order 2，HUST Monthly 2011.04.09 之 D，1434

reverse order 2，HUST Monthly 2011.04.09 之 D，1434

reverse order 2

Time Limit: 1 Sec Memory Limit: 128 MB
Submissions: 157 Solved: 64

Description

Here is a sequence a_1..n, which is a disordered sequence from 1 to N. if i < j and a_i > a_j, <i, j> is called a pair of inversion. And b_1..n-1 is defined as follows, bk is the number of the total inversion pairs in array a, when i<=k<j. Now the array a is required while the array b is known.

 

Input

 

Several cases end with the end of the file;

And each of the cases includes two lines, a integer n(2<=n<=10^5)in the first line, and the second line followed with n-1 integer, which is in the presentation of array b;

 

Output

Output the answer of each case in a line, namely the array a, and a space is required between adjacent integers.

 

Sample Input

5
2 1 2 0

Sample Output

3 1 4 2 5

a[ 1 ] = b[ 1 ] + 1;

求 b[ i ] 时，a[ i ] 左边比它大的有 X 个，a[ i ] 右边比它小的有 Y 个，
则比 a[ i ] 小的一共有 ( Y - X + i - 1 ) 个，所以 a[ i ] = Y - X + i，

即 a[ i ] = b[ i ] - b[ i - 1 ] + i。

用 int 的 b 会错误，要 long long 的 b 。

  
  
  
  

   
   
   
    1
   
   
   
   
   
   
   
   
   
   
   
   /**/
   
   
   
   /*
 2// int b, WA
 3#include <stdio.h>
 4#include <string.h>
 5
 6#define  L  100009
 7
 8int has[ L ];
 9
10int main() {
11        int n, i, a, b, bk;
12        while ( scanf( "%d", &n ) == 1 ) {
13                memset( has, 0, sizeof(has) );
14                bk = 0;
15                for ( i = 1; i < n; ++i ) {
16                        scanf( "%d", &b );
17                        a  = b - bk + i;
18                        bk = b;
19                        has[ a ] = 1;
20                        printf( "%d ", a );
21                }
22                for ( i = 1; i <= n; ++i ) {
23                        if( has[ i ] == 0 ) {
24                                a = i;
25                        }
26                }
27                printf( "%d\n", a );
28        }
29        return 0;
30}
31*/
   
   
   
   

   
   
   
   32
   
   
   
   

   
   
   
   33
   
   
   
   

   
   
   
   34
   
   
   
   
   
   
   
   //
   
   
   
    long long b, AC
   
   
   
   

   
   
   
   35
   
   
   
   
   
   
   
   #include 
   
   
   
   <
   
   
   
   stdio.h
   
   
   
   >
   
   
   
   

   
   
   
   36
   
   
   
   #include 
   
   
   
   <
   
   
   
   string
   
   
   
   .h
   
   
   
   >
   
   
   
   

   
   
   
   37
   
   
   
   

   
   
   
   38
   
   
   
   
   
   
   
   #define
   
   
   
     L  100009
   
   
   
   

   
   
   
   39
   
   
   
   

   
   
   
   40
   
   
   
   
   
   
   
   int
   
   
   
    has[ L ];

   
   
   
   41
   
   
   
   

   
   
   
   42
   
   
   
   
   
   
   
   int
   
   
   
    main() 
   
   
   
   
   
   
   
   {
43        int n, i, a;
44        long long bk, b;
45        while ( scanf( "%d", &n ) == 1 ) {
46                memset( has, 0, sizeof(has) );
47                bk = 0;
48                for ( i = 1; i < n; ++i ) {
49                        scanf( "%lld", &b );
50                        a  = b - bk + i;
51                        bk = b;
52                        has[ a ] = 1;
53                        printf( "%d ", a );
54                }
55                for ( i = 1; i <= n; ++i ) {
56                        if( has[ i ] == 0 ) {
57                                a = i;
58                        }
59                }
60                printf( "%d\n", a );
61        }
62        return 0;
63}
   
   
   
   

   
   
   
   64