Limited Insertion_女insertion-优快云博客

本文链接：https://blog.youkuaiyun.com/LinChenEric/article/details/95027581

本文探讨了一种名为有限插入序列的算法挑战，该算法要求在空序列中通过N次操作，每次选择一个整数并将其插入到特定位置，最终判断是否能够得到目标序列。文章详细解释了问题背景、约束条件、输入输出格式，并提供了一个C++实现示例，通过倒序遍历和暴力判断的方法来解决这个问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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
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)

思路

这题题目比较水，数据范围不大，就直接暴力倒叙判断即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=100;
struct code
{
    int num;
    int dis;
    int pos;
}a[N+5];
int b[N+5];
int c[N+5];
int main()
{
    int x;
    cin>>x;
    for(int i=1;i<=x;i++)
    {
        scanf("%d",&a[i].num);
        a[i].pos=i;
    }
    for(int i=1;i<=x;i++)
    {
        a[i].dis=0;
        /*printf("%d\n",a[i].pos-a[i].dis);*/
    }
    int n=0,flag;
    for(int i=0;i<x;i++)
    {
        flag=0;
        for(int j=x;j>0;j--)
        {
            /*cout<<a[j].pos-a[j].dis<<endl;
            cout<<a[j].num<<endl;*/
            if(a[j].num==a[j].pos-a[j].dis&&c[j]==0)
            {
                b[n++]=a[j].num;
                c[j]=1;
                flag=1;
                for(int k=j+1;k<=x;k++)
                {
                    a[k].dis++;
                }
                break;
            }
        }
        if(flag==0)
        {
            break;
        }
    }
    if(flag==0)
    {
        printf("-1\n");
    }
    else
    {
        for(int i=x-1;i>=0;i--)
        {
            printf("%d\n",b[i]);
        }
    }
    return 0;
}