POJ2805 Inversion水题

Posted 2021-09-04 海岛Blog

Inversion
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 4641 Accepted: 2034

Description

The inversion number of an integer sequence a1, a2, . . . , an is the number of pairs (ai, aj) that satisfy i < j and ai > aj . Given n and the inversion number m, your task is to find the smallest permutation of the set { 1, 2, . . . , n }, whose inversion number is exactly m.
A permutation a1, a2, . . . , an is smaller than b1, b2, . . . , bn if and only if there exists an integer k such that aj = bj for 1 <= j < k but ak < bk.

Input

The input consists of several test cases. Each line of the input contains two integers n and m. Both of the integers at the last line of the input is −1, which should not be processed. You may assume that 1 <= n <= 50000 and 0 <= m <= n(n − 1)/2.

Output

For each test case, print a line containing the smallest permutation as described above, separates the numbers by single spaces.

Sample Input

5 9
7 3
-1 -1

Sample Output

4 5 3 2 1
1 2 3 4 7 6 5

Source

Shanghai 2004 Preliminary

问题链接：POJ2805 Inversion
问题简述：（略）
问题分析：水题，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* POJ2805 Inversion */

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int n, m, end = 0;
    while (~scanf("%d%d", &n, &m) && (n + m != -2)) {
        int tot = 0;
        for (int i = n; i >= 1; i--) {
            tot += n - i;
            if (tot >= m) {
                end = i;
                break;
            }
        }

        int k = m + end - (n - end) * (n - end - 1) / 2;
        for (int i = 1; i < end; i++) printf("%d ", i);
        printf("%d ", k);
        for (int i = n; i >= end; i--)
            if (i != k) printf("%d ", i);
        printf("\\n");
    }

    return 0;
}