Description

You are given a permutation p of the set \({1,2,...,N}\). Please construct two sequences of positive integers \(a_1, a_2, ..., a_N\) and \(b_1, b_2, ..., b_N\) satisfying the following conditions:
·\(1≤a_i,b_i≤10^9\) for all \(i\)
·\(a_1<a_2<...<a_N\)
·\(b_1>b_2>...>b_N\)
·\(a_{p_1}+b_{p_1}<a_{p_2}+b_{p_2}<...<a_{p_N}+b_{p_N}\)

Constraints
·\(2≤N≤20,000\)
·\(p\) is a permutation of the set \({1,2,...,N}\)

Solution

注意到a，b分别是递增和递减的，不妨先按照等差数列构造，首先保证\(a_i+b_i\)不变，然后再按照给如的序列p在a序列的基础上减去一个相对较小的值，保证不会改变原有的单调性，这样就改变了\(a_i+b_i\)的单调性

Note

第一次选值的时候选取了20000，由于太大而影响了最总的结果，这里选择N然后令其递减即可

Code

#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

int p[20000 + 10];

int a[20000 + 10], b[20000 + 10];
int main() {
    //freopen("test.txt", "r", stdin);
    ios::sync_with_stdio(false);
    cin.tie(0);
    int N; 
    cin >> N;
    for (int i = 1; i <= N; i++) cin >> p[i];
    for (int i = 1; i <= N; i++) a[i] = i*20001;
    for (int i = N; i >= 1; i--) b[N-i+1] = i*20001;
    int base = N;
    for (int i = 1; i <= N; i++) {
        a[p[i]] -= base;
        base--;
    }
    for (int i = 1; i <= N; i++) cout << a[i] << " ";
    cout << endl;
    for (int i = 1; i <= N; i++) cout << b[i] << " ";
    return 0;
}