10.25 分组

 

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

#define N 131073

int n, m = 0, K;
int a[N], b[N];
bool vis[N], dvis[N], issqr[N * 2];
int f[N * 2];

int getf(int x) {return f[x] > 0 ? (f[x] = getf(f[x])) : x;}

void merge(int u, int v) {
    u = getf(u), v = getf(v);
    if (u != v) {
        if (f[u] > f[v]) swap(u, v);
        f[u] += f[v];
        f[v] = u;
    }
}

bool check(int u, int v) {
    int s1 = getf(u), s2 = getf(u + N);
    int t1 = getf(v), t2 = getf(v + N);
    if (s1 == t1) return 1;
    if (s2 == t2) return 1;
    merge(s1, t2); merge(s2, t1);
    return 0;
}

void solve_1() {
    for (int i = n, j = n; i;) {
        for (bool flag = 1; j; j--) {
            for (int k = 1; k * k - a[j] < N; k++) {
                if (k * k - a[j] <= 0) continue;
                if (vis[k * k - a[j]]) {flag = 0; break;} 
            }
            if (!flag) break;
            vis[a[j]] = 1;
        }
        if (!j) break;
        b[++m] = j;
        for ( ; i > j; i--) vis[a[i]] = 0;
    }
}

void solve_2() {
    memset(f, -1, sizeof f);
    for (int i = 1; i * i < 2 * N; i++) issqr[i * i] = 1;
    for (int i = n, j = n; i;) {
        for (bool flag = 1; j; j--) {
            if (vis[a[j]]) {
                if (issqr[a[j] + a[j]]) {
                    if (dvis[a[j]]) break;
                    for (int k = 1; k * k - a[j] < N; k++) {
                        if (k * k - a[j] <= 0) continue;
                        if (vis[k * k - a[j]] && k * k != a[j] * 2) {
                            flag = 0; break;
                        } 
                    }
                    if (!flag) break;
                    dvis[a[j]] = 1;
                }
            }
            else {
                for (int k = 1; k * k - a[j] < N; k++) {
                    if (k * k - a[j] <= 0) continue;
                    if (vis[k * k - a[j]]) {
                        if (check(k * k - a[j], a[j])) {flag = 0; break;}
                    } 
                }
                if (!flag) break;
                vis[a[j]] = 1;
            }
        }
        if (!j) break;
        b[++m] = j;
        for ( ; i > j; i--) f[a[i]] = f[a[i] + N] = -1, vis[a[i]] = 0, dvis[a[i]] = 0;
    }
}

int main() {
    freopen("division.in", "r", stdin);
    freopen("division.out", "w", stdout);
    scanf("%d%d", &n, &K);
    for (int i = 1; i <= n; i++) scanf("%d", a + i);
    if (K == 1) solve_1();
    else solve_2();
    printf("%d\n", m + 1);
    for (int i = m; i; i--) printf("%d ", b[i]);
    putchar('\n');
    return 0;
}

 

转载于:https://www.cnblogs.com/lyqlyq/p/7729666.html