问题 A: Limited Insertion

题目描述

Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1≤j≤i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to
b after N operations. If it is, show one possible sequence of operations that achieves it.

Constraints
·All values in input are integers.
·1≤N≤100
·1≤bi≤N

输入

Input is given from Standard Input in the following format:

N
b1 … bN

输出

If there is no sequence of N operations after which a would be equal to b, print -1. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.

样例输入
复制样例数据 3
1 2 1

样例输出
1
1
2

提示

In this sequence of operations, the sequence a changes as follows:
·After the first operation: (1)
·After the second operation: (1,1)
·After the third operation: (1,2,1)

反思：一开始想复杂了，比队友多卡了一个小时！没有想清楚再做，自己还是太菜了；

思路：最后的插入的数字一定在自己的位置上（a[i]==i)，把那个数字拿掉（数字前移），再从后往前推下一个数字同样在自己的位置上本来想的是顺序去找，找出在自己位置上的最大的那个数，结果用这个方法没写好哪里有bug，一直卡；其实从后面找到在自己位置上的一定就是最大的，那么就不用找最大的了；

#include <iostream>

using namespace std;

int main()
{
    int a[105]={0};
    int c[105]={0};
    int n;
    int cnt=0;
    int CNT=0;
    int sign=0;
    cin>>n;
    int flag=0;
    for(int i=1;i<=n;i++) cin>>a[i];
    //for(int i=1;i<=n;i++) cout<<a[i];
    for(int i=n;i>=1;i--)
    {
        flag=0;sign=0;
        for(int j=n;j>=1;j--){ if(a[j]==j) {sign=j;flag=1;break;} }
        if(flag==0){cout<<"-1"<<endl; return 0;}
        c[CNT++]=sign;
        for(int m=sign;m<=n;m++)///
        {
            a[m]=a[m+1];
            //cout<<a[m]<<' ';
        }
        //cout<<endl;
    }
    //cout<<CNT<<endl;
    for(int i=CNT-1;i>=0;i--)cout << c[i] << endl;
    //cout<<endl;
    //for(int i=1;i<=CNT;i++)cout << a[i] << endl;
    return 0;
}