D. Artsem and Saunders time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output
Artsem has a friend Saunders from University of Chicago. Saunders PResented him with the following problem.
Let [n] denote the set {1,?…,?n}. We will also write f:?[x]?→?[y] when a function f is defined in integer points 1, …, x, and all its values are integers from 1 to y.
Now then, you are given a function f:?[n]?→?[n]. Your task is to find a positive integer m, and two functions g:?[n]?→?[m], h:?[m]?→?[n], such that g(h(x))?=?x for all , and h(g(x))?=?f(x) for all , or determine that finding these is impossible. Input
The first line contains an integer n (1?≤?n?≤?105).
The second line contains n space-separated integers — values f(1),?…,?f(n) (1?≤?f(i)?≤?n). Output
If there is no answer, print one integer -1.
Otherwise, on the first line print the number m (1?≤?m?≤?106). On the second line print n numbers g(1),?…,?g(n). On the third line print m numbers h(1),?…,?h(m).
If there are several correct answers, you may output any of them. It is guaranteed that if a valid answer exists, then there is an answer satisfying the above restrictions. Examples Input
3 1 2 3
Output
3 1 2 3 1 2 3
Input
3 2 2 2
Output
1 1 1 1 2
Input
2 2 1
Output
-1
給出數組 a,是否可以構造出數組g,h滿足 :g(h(x))?=?x ,h(g(x))?=?f(x) 數組h不重復記錄數組a里的元素,數組g(x)記錄a[x]在h數組里的位置,但若a[x] != x,便無法構成
AC代碼:
#include<cstdio>#include<algorithm>using namespace std;const int K = 1e5 + 10;int a[K],b[K],c[K],ok[K];int main(){ int N; scanf("%d",&N); for(int i = 1; i <= N; i++) scanf("%d",&a[i]),ok[a[i]] = 1; int nl = 0; for(int i = K - 1; i >= 1; i--) if(ok[i]){ c[++nl] = i,b[i] = nl; if(a[i] != i){printf("-1/n"); return 0;} } printf("%d/n",nl); for(int i = 1; i <= N; i++) printf("%d ",b[a[i]]);printf("/n"); for(int i = 1; i <= nl; i++) printf("%d ",c[i]); return 0;}新聞熱點
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