Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 400400 points
Problem Statement
Snuke has an empty sequence aa.
He will perform NN operations on this sequence.
In the ii-th operation, he chooses an integer jj satisfying 1≤j≤i1≤j≤i, and insert jj at position jj in aa (the beginning is position 11).
You are given a sequence bb of length NN. Determine if it is possible that aa is equal to bb after NN operations. If it is, show one possible sequence of operations that achieves it.
Constraints
- All values in input are integers.
- 1≤N≤1001≤N≤100
- 1≤bi≤N1≤bi≤N
Input
Input is given from Standard Input in the following format:
NN
b1b1 …… bNbN
Output
If there is no sequence of NN operations after which aa would be equal to bb, print -1. If there is, print NN lines. In the ii-th line, the integer chosen in the ii-th operation should be printed. If there are multiple solutions, any of them is accepted.
Sample Input 1 Copy
3
1 2 1
Sample Output 1 Copy
1
1
2
In this sequence of operations, the sequence aa changes as follows:
- After the first operation: (1)(1)
- After the second operation: (1,1)(1,1)
- After the third operation: (1,2,1)(1,2,1)
Sample Input 2 Copy
2
2 2
Sample Output 2 Copy
-1
22 cannot be inserted at the beginning of the sequence, so this is impossible.
Sample Input 3 Copy
9
1 1 1 2 2 1 2 3 2
Sample Output 3 Copy
1
2
2
3
1
2
2
1
1
题意：
初始你有一个空的数组，
你将执行以下操作n次，
第i次你可以选择一个1~i的数，
并把这个数插入数组的第i个位置，之前的i和i的位置如果有数将向后移动。
现在给你最后的结果数组，让你判断是否可以通过操作来完成，如果可以请输出一个方案。
思路：
我们可以用逆向思维，我们知道这n个操作的最后一个操作一定是把i放在i的位置，那么我们不妨从大到小枚举数组的a[i] 是否等于 i
如果等于，我们可以把它作为我们的最后一次操作，然后把这个数从数组中删除，然后再重复上面的操作，来找次最后的操作。。
如果某一步找不到一个a[i]==i时，那么可以得出没有方案得到这个数组。
如果都可以删除掉，然后把中途找到的数i，逆序输出，就说我们要输出的答案了。
细节见代码：
#include <iostream>
#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=;while(b){if(b%)ans=ans*a%MOD;a=a*a%MOD;b/=;}return ans;}
inline void getInt(int* p);
const int maxn=;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
int a[maxn];
int main()
{
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
// list<int> ls;
// int n;
gbtb;
cin>>n;
repd(i,,n)
{
cin>>a[i];
}
int isok=;
std::vector<int> ans;
int len=n;
while(len)
{
int temp=len;
for(int i=len;i>=;i--)
{
if(a[i]==i)
{
ans.push_back(i);
repd(j,i,len)
{
a[j]=a[j+];
}
len--;
break;
}
}
if(temp==len)
{
break;
}
}
if(len>)
{
cout<<-<<endl;
}else
{
reverse(ALL(ans));
for(auto x:ans)
{
cout<<x<<endl;
}
} return ;
} inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '');
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * - ch + '';
}
}
else {
*p = ch - '';
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * + ch - '';
}
}
}