Lost Cows
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9660 | Accepted: 6219 |
Description
N (2 <= N <= 8,000) cows have unique brands in the range 1..N. In a spectacular display of poor judgment, they visited the neighborhood ‘watering hole‘ and drank a few too many beers before dinner. When it was time to line up for their evening meal, they did not line up in the required ascending numerical order of their brands.
Regrettably, FJ does not have a way to sort them. Furthermore, he‘s not very good at observing problems. Instead of writing down each cow‘s brand, he determined a rather silly statistic: For each cow in line, he knows the number of cows that precede that cow in line that do, in fact, have smaller brands than that cow.
Given this data, tell FJ the exact ordering of the cows.
Input
* Line 1: A single integer, N
* Lines 2..N: These N-1 lines describe the number of cows that precede a given cow in line and have brands smaller than that cow. Of course, no cows precede the first cow in line, so she is not listed. Line 2 of the input describes the number of preceding cows whose brands are smaller than the cow in slot #2; line 3 describes the number of preceding cows whose brands are smaller than the cow in slot #3; and so on.
Output
* Lines 1..N: Each of the N lines of output tells the brand of a cow in line. Line #1 of the output tells the brand of the first cow in line; line 2 tells the brand of the second cow; and so on.
Sample Input
5 1 2 1 0
Sample Output
2 4 5 3 1
Source
题目大意：
就是说，让你求出一个１到ｎ的排列，但是，一开始，这个排列只告诉了你从第二个数字开始的比这个数字小的数的个数。
解题思路：
刚开始想了好久，还是自己太弱，对于数学方面的问题还应该多做题，ＣＱ省队爷告诉我，应该先由一般的规律推出来如果你要确定某个数字ｐ，那么我们就要知道ｐ的前面有多少个数字比它小，记为ｍ，ｐ的后面有多少个数字比他小，记为ｎ，那么ｐ的位置就该在这个数列中的第ｍ＋ｎ＋１处，我们就二分这个ｐ，拿每次得到的ｍｉｄ来和ｍ＋ｎ＋１来比较大小，如果ｍ＋ｎ＋１＞ｍｉｄ就说明ｍｉｄ取小了，这个时候，我们就把left = mid+1;如果m+n+1<mid，那就说明mid取大了，这个时候我们就要把mid = right.
关于一个数字后面比他小的数字的个数怎么求解的问题，其实可以用树状数组来搞定。
关于一个数字前面有多少个数字比他小的问题，我们再输入的过程中已经知道了。
代码：
1 # include<cstdio>
2 # include<iostream>
3 # include<fstream>
4 # include<algorithm>
5 # include<functional>
6 # include<cstring>
7 # include<string>
8 # include<cstdlib>
9 # include<iomanip>
10 # include<numeric>
11 # include<cctype>
12 # include<cmath>
13 # include<ctime>
14 # include<queue>
15 # include<stack>
16 # include<list>
17 # include<set>
18 # include<map>
19
20 using namespace std;
21
22 const double PI=4.0*atan(1.0);
23
24 typedef long long LL;
25 typedef unsigned long long ULL;
26
27 # define inf 999999999
28 # define MAX 8000+4
29
30 int a[MAX];
31 int ans[MAX];
32 int tree[MAX];
33
34 int n;
35
36 int read ( int pos )
37 {
38 int ans = 0;
39 while ( pos > 0 )
40 {
41 ans+=tree[pos];
42 pos-=pos&(-pos);
43 }
44 return ans;
45 }
46
47 void update ( int pos,int val )
48 {
49 while ( pos <= n )
50 {
51 tree[pos]+=val;
52 pos+=pos&(-pos);
53 }
54 }
55
56 int my_search ( int k )
57 {
58 int left = 1, right = n;
59 while ( left < right )
60 {
61 int mid = ( left+right )>>1;
62 int num = read (mid);
63 if ( mid - 1 < num+k )
64 {
65 left = mid+1;
66 }
67 else
68 {
69 right = mid;
70 }
71 }
72 return left;
73 }
74
75 int main(void)
76 {
77 while ( scanf("%d",&n)!=EOF )
78 {
79 memset(tree,0,sizeof(tree));
80 memset(ans,0,sizeof(ans));
81 a[1] = 0;
82 for ( int i = 2;i <= n;i++ )
83 scanf("%d",&a[i]);
84 for ( int i = n;i >= 1;i-- )
85 {
86 int k = my_search(a[i]);
87 update(k,1);
88 ans[i] = k;
89 }
90
91 for ( int i = 1;i <= n;i++ )
92 printf("%d\n",ans[i]);
93 }
94
95
96
97 return 0;
98 }