Codeforces Round #105 (Div. 2) 148C Terse princess(脑洞)_terse princess codeforces

«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.

题目链接:点击打开链接

有一个长度为n的序列, 如果一个数比前面的数都大, 则会说Oh, 如果一个数比前面所有数的和都大, 则会说Wow. 现给出Oh数a, Wow数

b, 要求输出一个满足题意的数组, 不存在这样的数组则输出-1.

构造一个数组, a[1] = 1, 进行n - 2次操作, 每一次操作都记录当前值和前面数的和, 优先减少b, 为sum + 1, 否则减少a, 为res + 1.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 105;
int n, a, b, ans[MAXN];
int main(int argc, char const *argv[])
{
	scanf("%d%d%d", &n, &a, &b);
	ans[1] = 1;
	int res = 1, sum = 1;
	for(int i = 2; i <= n; ++i) {
		if(b) {
			res = sum + 1;
			b--;
		}
		else if(a && i > 2) {
			res++;
			a--;
		}
		ans[i] = res;
		sum += res;
	}
	if(a || b) {
		printf("-1\n");
		return 0;
	}
	for(int i = 1; i < n; ++i)
		printf("%d ", ans[i]);
	printf("%d\n", ans[n]);
	return 0;
}