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
给出数组 a,是否可以构造出数组g,h满足 :g(h(x)) = x ,h(g(x)) = f(x) 数组h不重复记录数组a里的元素,数组g(x)记录a[x]在h数组里的位置,但若a[x] != x,便无法构成
AC代码:
#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;}新闻热点
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