Codeforces Round #463 (Div. 1 + Div. 2, combined) C. Permutation Cycle （扩展欧几里得）

C. Permutation Cycle

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

For a permutation P[1... N] of integers from 1 to N, function f is defined as follows:

Let g(i) be the minimum positive integer j such that f(i, j) = i. We can show such j always exists.

For given N, A, B, find a permutation P of integers from 1 to N such that for 1 ≤ i ≤ N, g(i) equals either A or B.

Input

The only line contains three integers N, A, B (1 ≤ N ≤ 106, 1 ≤ A, B ≤ N).

Output

If no such permutation exists, output -1. Otherwise, output a permutation of integers from 1 to N.

Examples

input

Copy

9 2 5

output

6 5 8 3 4 1 9 2 7

input

Copy

3 2 1

output

1 2 3 

Note

In the first example, g(1) = g(6) = g(7) = g(9) = 2 and g(2) = g(3) = g(4) = g(5) = g(8) = 5

In the second example, g(1) = g(2) = g(3) = 1

（一）何谓Permutation Cycle

以例1中的6 5 8 3 4 1 9 2 7
第一个数是6，以6为位置，则第六个数是1，以1为位置，第一个数是6，以6为位置，第六个数是1……，这样6和1就构成了一个圈子。f(1) = 6, f(6) = 1，最短的周期是2

第二个数是5，以5为位置，第五个数是4，以4为位置，第四个数是3，以3为位置，第三个数是8，以8为位置，第八个数是2，以2为位置，第2个数是5，以5为位置，第五个数是4……这样5，4，3，8，2也构成了一个圈子。f(5) = 4, f(4) = 3, f(3) = 8, f(8) = 2, f(2) = 5，最短的周期是5。

（二）例子分析

对于例1中的6 5 8 3 4 1 9 2 7
p[1] = 6, p[2] = 5, p[3] = 8, p[4] = 3, p[5] = 4, p[6] = 1, p[7] = 9, p[8] = 2, p[9] = 7
f(1, j) = f(6, j-1) = p[6] = 1，此时j - 1 = 1 ==> j = 2 ==> g(1) = 2
f(2, j) = f(5, j-1) = f(4, j-2) = f(3, j-3) = f(8, j-4) = p[8] = 2，此时j - 4 = 1 ==> j = 5 ==> g(2) = 5
f(3, j) = f(8, j-1) = f(2, j-2) = f(5, j-3) = f(4, j-4) = p[4] = 3，此时j - 4 = 1 ==> j = 5 ==> g(3) = 5
……
所以，g(1) = g(6) = g(7) = g(9) =2, g(2) = g(3) = g(4) = g(5) = g(8) = 5

对于例2中的1，2，3
p[1] = 1, p[2] = 2, p[3] = 3
f(1, j) = p[1] = 1 ==> j = 1 ==> g(1) = 1
f(2, j) = p[2] = 2 ==> j = 1 ==> g(2) = 1
f(3, j) = p[3] = 3 ==> j = 1 ==> g(3) = 1
所以，g(1) = g(2) = g(3) = 1

（三）思路

对于本题来说，实际上就是求 Ax + By = N，x >= 0, y >= 0
对于例1，A = 2, B = 5, N = 9，Ax + By = N ==> 2x + 5y = 9 ==> x = 2, y = 1
也就是说，9个数里，有两组2个数，使得g(i) = 2。这两组数分别为2, 1和4, 3
有一组5个数，使得g(i) = 5，这组数为6,7,8,9,5

再举一例，A = 3, B = 6, N = 9
3x + 6y = 9 ==> x = 1, y = 1
也就是说，9个数里，有一组3个数，使得g(i) = 3。这组数为2，3，1
另有一组6个数，使得g(i) = 6。这组数为5，6，7，8，9，4

再举一例，A = 4, B = 6, N = 9
4x + 6y = 9，无解

题目思路来源：

链接：https://www.jianshu.com/p/c14d90b9db75

#include<cstdio>

int main()
{
    int n, a, b;
    int cnta, cntb;
    int flag = false;
    scanf("%d %d %d", &n, &a, &b);

    for (int i = 0; i * a <= n; i++)
    {
        int t = n - a * i;
        if (t % b == 0)
        {
            cnta = i;
            cntb = t / b;
            flag = true;
            break;
        }
    }

    int num = 1;
    int lastNum = num;

    if(!flag)
    {
         printf("-1
");
    }
    else
    {
        for(int i = 1;i <= cnta;i++)
        {
            for(int k = 1;k < a;k++)
            {
                printf("%d ",++num);
            }

            printf("%d ",lastNum);

            num++;
            lastNum = num;
        }

        for(int i = 1;i <= cntb;i++)
        {
            for(int k = 1;k < b;k++)
            {
                printf("%d ",++num);
            }

            printf("%d ",lastNum);

            num++;
            lastNum = num;
        }
    }

    return 0;
}