分析：暴力分挺多，也挺好想的，個人感覺兩個特殊性質沒什麽卵用.

對於K=1，n ≤ 1024的情況，從後往前貪心地分，如果能和上一組分在一起就分在一起，否則就再開一組，這樣可以保證字典序最小.ai ≤ 2就看前面有沒有2.有就不能分在一組.n ≤ 131072就不能再這樣二重循環枚舉了，因為兩個數的和頂多只有262114 = 512^2,從1枚舉到512，看看它的平方有沒有被占用就可以了，就把問題從序列上轉化到了值域上.

對於K=2,其實做法和K=1沒什麽兩樣，只是每一組可以分成兩個對立的小組，非常像noip2010關押罪犯，加一個並查集就好了.用一個vector存每個值的兔子的位置，判斷一下有沒有沖突就好了.

暴力+正解：

#include <cmath>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 131072;

vector <int>E[maxn * 2 + 10];
int n, k, a[150000], ans[150000], cnt, fa[300010], vis[300010], T;
bool flag = true
 
;

bool check2(int x)
{
    for (int i = 2; i <= 512; i++)
    {
        if (i * i - x >= 0 && vis[i * i - x])
            return false;
    }
    return true;
}

int find(int x)
{
    if (x == fa[x])
        return x;
    return fa[x] = find(fa[x]);
}

bool check(int x, int y)
{
    int
 
 t = x + y;
    int p = sqrt(t);
    if (p * p == t)
        return false;
    return true;
}

void solve1()
{
    if (n <= 1024)
    {
        int cur = n;
        for (int i = n; i >= 1; i--)
        {
            for (int j = i + 1; j <= cur; j++)
            {
                if (!check(a[i], a[j]))
                {
                    ans[++cnt] = i;
                    cur = i;
                }
            }
        }
        printf("%d\n", cnt + 1);
        for (int i = cnt; i >= 1; i--)
            printf("%d ", ans[i]);
        printf("\n");
    }
    else
        if (flag)
        {
        int cur = n;
        bool flag2 = false;
        for (int i = n; i >= 1; i--)
        {
            if (flag2 && a[i] == 2)
            {
                ans[++cnt] = i;
                cur = i;
            }
            if (a[i] == 2)
                flag2 = 1;
        }
        printf("%d\n", cnt + 1);
        for (int i = cnt; i >= 1; i--)
            printf("%d ", ans[i]);
        printf("\n");
        }
        else
        {
            int cur = n;
            for (int i = n; i >= 1; i--)
            {
                if (!check2(a[i]))
                {
                    for (int j = i + 1; j <= cur; j++)
                        vis[a[j]] = 0;
                    ans[++cnt] = i;
                    cur = i;
                }
                vis[a[i]] = 1;
            }
            printf("%d\n", cnt + 1);
            for (int i = cnt; i >= 1; i--)
                printf("%d ", ans[i]);
            printf("\n");
        }
}

void hebing(int x, int y)
{
    x = find(x), y = find(y);
    fa[x] = y;
}

bool check3(int x)
{
    for (int i = 1; i <= 512; i++)
    {
        if (i * i - a[x] >= 0 && vis[i * i - a[x]] == T)
        {
            int now = i * i - a[x];
            for (int j = 0; j < E[now].size(); j++)
            {
                if (find(x) == find(E[now][j]))
                    return false;
                hebing(x + n, E[now][j]);
                hebing(x, E[now][j] + n);
            }
        }
    }
    return true;
}

void solve2()
{
    if (flag)
    {
        int flag2 = 0;
        for (int i = n; i >= 1; i--)
        {
            if (flag2 == 2 && a[i] == 2)
            {
                ans[++cnt] = i;
                flag2 = 0;
            }
            if (a[i] == 2)
                flag2++;
        }
        printf("%d\n", cnt + 1);
        for (int i = cnt; i >= 1; i--)
            printf("%d ", ans[i]);
        printf("\n");
    }
    else
        if (n <= 1024)
        {
        for (int i = 1; i <= n * 2; i++)
            fa[i] = i;
        int cur = n;
        for (int i = n; i >= 1; i--)
        {
            for (int j = i + 1; j <= cur; j++)
            {
                if (!check(a[i], a[j]))
                {
                    if (find(i) == find(j))
                    {
                        ans[++cnt] = i;
                        cur = i;
                    }
                    else
                    {
                        int xx = find(i + n), yy = find(j + n);
                        fa[i] = yy;
                        fa[j] = xx;
                    }
                }
            }
        }
        printf("%d\n", cnt + 1);
        for (int i = cnt; i >= 1; i--)
            printf("%d ", ans[i]);
        printf("\n");
        }
        else
        {
            T = 0;
            for (int i = 1; i <= n * 2; i++)
                fa[i] = i;
            for (int i = n; i >= 1; i--)
            {
                if (!check3(i))
                {
                    T++;
                    ans[++cnt] = i;
                }
                    if (vis[a[i]] != T)
                    {
                        vis[a[i]] = T;
                        E[a[i]].clear();
                    }
                    E[a[i]].push_back(i);
            }
            printf("%d\n", cnt + 1);
            for (int i = cnt; i >= 1; i--)
                printf("%d ", ans[i]);
            printf("\n");
        }
}

int main()
{
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        if (a[i] > 2)
            flag = false;
    }

    if (k == 1)
        solve1();
    if (k == 2)
        solve2();

    return 0;
}

noip模擬賽 分組