题解 分组
考试最后5min想到了k=1时的骗分方案，可惜没时间写了……
这题代码写得有点恶心了，数都很小，完全没必要开4个hush表
k=1时要求同一个组中不能有冲突元素，这里如果枚举组中元素判冲突就是\(n^2\)了
发现两元素冲突当且仅当它们加和为某个正整数\(a\)的平方，
考虑不枚举元素，而是通过枚举这个\(a\)判冲突
那时间复杂度能开个根号
这里可以推广，如果有题要求在一个集合中找两个数，使它们运算后的值为某些特定数，
则如果特定数的数量较少，可以考虑「对于集合中的每个数，枚举所有特定数，并检查逆运算结果是否在集合中」的方式避免\(n^2\)的判断
k=2时允许每个组内再乱序划分，
这原来是个二分图判定问题啊，有空看看二分图都有点什么性质（咕）
当年关押罪犯的坑在这里填了：
「敌对并查集」其实就是维护点之间的敌对关系，如果这些关系刚好能塞进一张二分图里，则这些点能被分进两个集合，否则不能
然后如果有重复元素，注意特判一下\(2x=a^2\) 和 \(2x=a_1^2\)且\(x+y=a_2^2\)的情况就好
严重写麻烦了的 Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 140010
#define ll long long
#define ld long double
#define usd unsigned
#define ull unsigned long long
//#define int long long
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
char buf[1<<21], *p1=buf, *p2=buf;
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n, k, m=1;
int a[N], sta[N], top;
namespace task1{
struct hash_map{
static const int SIZE=13131;
int head[SIZE], size;
//int cnt[SIZE];
hash_map():size(0){memset(head, 0, sizeof(head));}
struct node{int dat, next;}e[N];
inline bool operator [] (int q) {
int t=q%SIZE;
for (int i=head[t]; i; i=e[i].next)
if (e[i].dat==q) return 1;
return 0;
}
inline void add(int q) {
int t=q%SIZE;
//++cnt[t];
node* k=&e[++size]; k->dat=q; k->next=head[t]; head[t]=size;
}
inline void clear() {size=0; memset(head, 0, sizeof(head));}
//void out() {int k=0; for (int i=1; i<=SIZE; ++i) cout<<cnt[i]<<endl, k=max(k, cnt[i]); cout<<"max: "<<k<<endl;}
}mp;
void solve() {
for (int i=n,t; i; --i) {
for (int j=512; j; --j)
if (t=j*j, t>a[i]) {
if (mp[t-a[i]]) {
++m;
sta[++top]=i;
mp.clear();
break;
}
}
else break;
mp.add(a[i]);
}
printf("%d\n", m);
while (top) printf("%d ", sta[top--]);
printf("\n");
}
}
namespace task2{
int fa[N<<1];
inline int find(int p) {return fa[p]==p?p:fa[p]=find(fa[p]);}
struct hash_map1{
static const int SIZE=13131;
int head[SIZE], size;
//int cnt[SIZE];
hash_map1():size(0){memset(head, 0, sizeof(head));}
struct node{int dat, rank, next;}e[N];
inline bool operator [] (int q) {
int t=q%SIZE;
for (int i=head[t]; i; i=e[i].next)
if (e[i].dat==q) return 1;
return 0;
}
inline void add(int r, int q) {
int t=q%SIZE;
//++cnt[t];
node* k=&e[++size]; k->dat=q; k->rank=r; k->next=head[t]; head[t]=size;
}
inline void clear() {size=0; memset(head, 0, sizeof(head));}
//void out() {int k=0; for (int i=1; i<=SIZE; ++i) k=max(k, cnt[i]); cout<<"max: "<<k<<endl;}
}mp;
struct hash_map2{
static const int SIZE=557;
int head[SIZE], size;
//int cnt[SIZE];
hash_map2():size(0){memset(head, 0, sizeof(head));}
struct node{int dat, cnt, next;}e[N];
inline bool operator [] (int q) {
int t=q%SIZE;
for (int i=head[t]; i; i=e[i].next)
if (e[i].dat==q)
if (e[i].cnt>=2) return 1;
else return 0;
return 0;
}
inline void add(int q) {
int t=q%SIZE;
//++cnt[t];
for (int i=head[t]; i; i=e[i].next)
if (e[i].dat==q) {
++e[i].cnt;
return ;
}
node* k=&e[++size]; k->dat=q; k->cnt=1; k->next=head[t]; head[t]=size;
}
inline void clear() {size=0; memset(head, 0, sizeof(head));}
//void out() {int k=0; for (int i=1; i<=SIZE; ++i) k=max(k, cnt[i]); cout<<"max: "<<k<<endl;}
}mp2;
struct hash_map3{
static const int SIZE=557;
int head[SIZE], size;
//int cnt[SIZE];
hash_map3():size(0){memset(head, 0, sizeof(head));}
struct node{int dat, next;}e[N];
inline bool operator [] (int q) {
int t=q%SIZE;
for (int i=head[t]; i; i=e[i].next)
if (e[i].dat==q) return 1;
return 0;
}
inline void add(int q) {
int t=q%SIZE;
//++cnt[t];
node* k=&e[++size]; k->dat=q; k->next=head[t]; head[t]=size;
}
//void out() {int k=0; for (int i=1; i<=SIZE; ++i) k=max(k, cnt[i]); cout<<"max: "<<k<<endl;}
}mp3;
void solve() {
for (int i=1,lim=n*2; i<=lim; ++i) fa[i]=i;
for (int i=1; i<=512; ++i) mp3.add(i*i);
for (int i=n,t,t2,f1,f2; i; --i) {
//cout<<i<<endl;
if (mp3[a[i]*2]&&mp2[a[i]]) {
++m;
sta[++top]=i;
mp.clear();
mp2.clear();
fa[i]=i; fa[i+n]=i+n;
mp.add(i, a[i]);
mp2.add(a[i]);
continue;
}
for (int j=512; j; --j)
if (t=j*j, t>a[i]) {
int q=t-a[i];
//if (q==a[i]) continue;
int p=(q)%mp.SIZE;
for (int l=mp.head[p]; l; l=mp.e[l].next)
if (mp.e[l].dat==q) {
f1=find(i); f2=find(mp.e[l].rank);
//cout<<"rank: "<<mp.e[l].rank<<endl;
//cout<<i<<endl;
//cout<<f1<<' '<<f2<<endl;
fa[find(i+n)]=f2;
fa[find(mp.e[l].rank+n)]=f1;
//cout<<"judge: "<<f1<<' '<<f2<<endl;
if (find(i)==find(i+n)) {
//cout<<"pos1"<<endl;
++m;
sta[++top]=i;
mp.clear();
mp2.clear();
fa[i]=i; fa[i+n]=i+n;
goto jump;
}
}
}
else break;
jump:
mp.add(i, a[i]);
if (mp3[a[i]*2]) mp2.add(a[i]); //cout<<"add2 "<<a[i]<<endl;
}
printf("%d\n", m);
while (top) printf("%d ", sta[top--]);
printf("\n");
//cout<<mp3[131072*2]<<' '<<mp2[131072]<<endl;
//cout<<a[2657]<<endl;
//mp.out(); mp2.out(); mp3.out();
}
}
signed main()
{
#ifdef DEBUG
freopen("1.in", "r", stdin);
#endif
n=read(); k=read();
for (int i=1; i<=n; ++i) a[i]=read();
if (n==1) {printf("1\n\n"); return 0;}
if (k&1) task1::solve();
else task2::solve();
return 0;
}
