CodeForces 148C - Terse princess

最新推荐文章于 2020-03-29 20:49:57 发布

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本文介绍了一道关于公主择婿的算法题目，公主通过特定的标准评价每位求婚者的财富，并据此发出不同的赞叹。文章详细解释了如何根据公主赞叹的次数逆向推导每位求婚者的财富值，同时提供了一个实现该逻辑的C++代码示例。

«Next please», — the princess called and cast an estimating glance at the next groom.

The princess intends to choose the most worthy groom, this is, the richest one. Whenever she sees a groom who is more rich than each of the previous ones, she says a measured «Oh...». Whenever the groom is richer than all previous ones added together, she exclaims «Wow!» (no «Oh...» in this case). At the sight of the first groom the princess stays calm and says nothing.

The fortune of each groom is described with an integer between 1 and 50000. You know that during the day the princess saw n grooms, said «Oh...» exactly a times and exclaimed «Wow!» exactly b times. Your task is to output a sequence of n integers t1, t2, ..., tn, where tidescribes the fortune of i-th groom. If several sequences are possible, output any of them. If no sequence exists that would satisfy all the requirements, output a single number -1.

Input

The only line of input data contains three integer numbers n, a and b (1 ≤ n ≤ 100, 0 ≤ a, b ≤ 15, n > a + b), separated with single spaces.

Output

Output any sequence of integers t1, t2, ..., tn, where ti (1 ≤ ti ≤ 50000) is the fortune of i-th groom, that satisfies the given constraints. If no sequence exists that would satisfy all the requirements, output a single number -1.

Sample test(s)

input
10 2 3

output
5 1 3 6 16 35 46 4 200 99

input
5 0 0

output
10 10 6 6 5

Note

Let's have a closer look at the answer for the first sample test.

The princess said «Oh...» (highlighted in bold): 5 1 3 6 16 35 46 4 200 99.
The princess exclaimed «Wow!» (highlighted in bold): 5 1 3 6 16 35 46 4 200 99.

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这题坑很多……WA了几次看了数据才改过来……主要就是-1出现的条件

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

int a[111111];

int main()
{
    int n,a,b,s[111]={0};
    cin>>n>>a>>b;
    if(n-a==1&&!b&&a)
    {
        cout<<"-1"<<endl;
        return 0;
    }
    s[0]=1;
    int sum=1;
    int i;
    for(i=1;i<=b;i++)
    {
        s[i]=sum+1;
        sum+=s[i];
    }
    if(b==0)
    {
        s[1]=1;
        i++;
        b=1;
    }
    for(;i<=a+b;i++)
    {
        s[i]=s[i-1]+1;
    }
    for(;i<=n;i++)
    {
        s[i]=1;
    }
    for(int i=0;i<n;i++)
        cout<<s[i]<<" ";
    cout<<endl;
    return 0;
}