题目链接：传送门

Solution

实际上就是求Ax + By = N的非负整数解（x和y都要非负），我是写的扩欧求的，实际上只需要暴力一个一个试就可以了。 然后写扩欧的话要注意在用特解求非负整数解的时候要注意A和B都要除一个gcd 考试的时候没加，然后就愉快地FST，rating暴跌了

Code

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#include <bits/stdc++.h>
#define int long long

using namespace std;

int N, A, B;
int Ans[1000000 + 100];

inline int ex_gcd (int a, int b, int &x, int &y)
{
    if (!b)
    {
        x = 1;
        y = 0;
        return a;
    }
    int gcd = ex_gcd (b, a % b, x, y);
    int tmp = x;
    x = y;
    y = tmp - a / b * y;
    return gcd;
}

main()
{
#ifdef hk_cnyali
    freopen("C.in", "r", stdin);
    freopen("C.out", "w", stdout);
#endif
    scanf("%lld%lld%lld", &N, &A, &B);
    int x, y;
    int g = ex_gcd(A, B, x, y);
    if (N % g)
    {
        cout<<-1<<endl;
        return 0;
    }
    B /= g;
    int t1 = ((N / g) * x % B + B) % B;
    B *= g;
    int p1 = (N - A * t1) / B;
    A /= g;
    int p2 = ((N / g) * y % A + A) % A;
    A *= g;
    int t2 = (N - B * p2) / A;
    if ((p1 < 0) || (t2 < 0))
    {
        cout<<-1<<endl;
        return 0;
    }
    if (p2 >= 0 && t2 >= 0) p1 = p2, t1 = t2;
    swap(p1, t1);
    for (int i = 0; i < p1; ++i)
    {
        Ans[i * A + 1] = i * A + A;
        for (int j = 2; j <= A; ++j)
            Ans[i * A + j] = i * A + j - 1;
    }
    for (int i = 0; i < t1; ++i)
    {
        Ans[A * p1 + i * B + 1] = A * p1 + i * B + B;
        for (int j = 2; j <= B; ++j)
            Ans[A * p1 + i * B + j] = A * p1 + i * B + j - 1;
    }
    for (int i = 1; i <= N; ++i) cout<<Ans[i]<<" ";
    cout<<endl;
    return 0;
}