AtCoder Grand Contest 032 A - Limited Insertion（ 思维）

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $400400$ points

Problem Statement

Snuke has an empty sequence $\mathrm{aa}$.

He will perform $\mathrm{NN}$ operations on this sequence.

In the $\mathrm{ii}$-th operation, he chooses an integer $\mathrm{jj}$ satisfying $\mathrm{1\le j\le i1\le j\le i}$, and insert $\mathrm{jj}$ at position $\mathrm{jj}$ in $\mathrm{aa}$ (the beginning is position $11$).

You are given a sequence $\mathrm{bb}$ of length $\mathrm{NN}$. Determine if it is possible that $\mathrm{aa}$ is equal to $\mathrm{bb}$ after $\mathrm{NN}$ operations. If it is, show one possible sequence of operations that achieves it.

Constraints

• All values in input are integers.
• $\mathrm{1\le N\le 1001\le N\le 100}$
• $\mathrm{1\le bi\le N1\le bi\le N}$

---

Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$
${\mathrm{b1b1}}^{}$ $\dots \dots$ ${\mathrm{bNbN}}^{}$

Output

If there is no sequence of $\mathrm{NN}$ operations after which $\mathrm{aa}$ would be equal to $\mathrm{bb}$, print -1. If there is, print $\mathrm{NN}$ lines. In the $\mathrm{ii}$-th line, the integer chosen in the $\mathrm{ii}$-th operation should be printed. If there are multiple solutions, any of them is accepted.

---

Sample Input 1 Copy

Copy

3
1 2 1

Sample Output 1 Copy

Copy

1
1
2

In this sequence of operations, the sequence $\mathrm{aa}$ 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

Copy

2
2 2

Sample Output 2 Copy

Copy

-1

$22$ cannot be inserted at the beginning of the sequence, so this is impossible.

---

Sample Input 3 Copy

Copy

9
1 1 1 2 2 1 2 3 2

Sample Output 3 Copy

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=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
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,1,n)
    {
        cin>>a[i];
    }
    int isok=0;
    std::vector<int> ans;
    int len=n;
    while(len)
    {
        int temp=len;
        for(int i=len;i>=1;i--)
        {
            if(a[i]==i)
            {
                ans.push_back(i);
                repd(j,i,len)
                {
                    a[j]=a[j+1];
                }
                len--;
                break;
            }
        }
        if(temp==len)
        {
            break;
        }
    }
    if(len>0)
    {
        cout<<-1<<endl;
    }else
    {
        reverse(ALL(ans));
        for(auto x:ans)
        {
            cout<<x<<endl;
        }
    }



    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}