C - Different Sums URAL - 2065 ----思维题

C - Different Sums

 URAL
- 2065 

Alex is a very serious mathematician and he likes to solve serious problems. For example, this problem.

You are to construct an array of n integers in which the amount of different integers is not less than k. Among all such arrays you have to construct the one with the minimal amount of different sums on non-empty
subarrays. In other words, lets compute the sums of every non-empty subarray and remove repeating sums. You have to minimize the number of remaining sums.

Input

In the only line of input there are two integers n, k (1 ≤ k ≤ n ≤ 500), separated by a space.

Output

Print n integers separated by spaces — the answer for the problem. All the numbers must not be greater than 10 6 by absolute value. It is guaranteed that there exists an optimal solution with numbers up to
10 5 by absolute value. If there are multiple possible answers, you may print any of them.

Example

input	output
1 1	-987654
3 2	0 7 0

Notes

Let’s take a closer look on the second sample. We will denote the sum on the segment [ l, r] by sum( l, r) (elements are numbered starting with 1). sum(1, 1) =sum(3,
3) = 0, sum(1, 2) = sum(1, 3) = sum(2, 2) = sum(2, 3) = 7, so there are only two different sums.

题目链接：https://cn.vjudge.net/contest/160744#problem/C

我和UMR在比赛的时候竟然忘记了负数。。。mdzz

题目的意思是说让你构造一个序列，使得子序列的和的个数最小，序列中不同的数的个数不小于k个。

最后出来的序列0，1，-1,2，-2.....

代码：

#include <bits/stdc++.h>
#define zero(x) (((x)>0?(x):(-x))<eps)
using namespace std;
const int MAXN=500+7;
int n,k;
int num[MAXN];
int main(){
scanf("%d%d",&n,&k);
int cnt=0;
for(int i=1;;++i){
if(cnt==k-1)break;
num[cnt++]=i;
if(cnt==k-1)break;
num[cnt++]=-i;
if(cnt==k-1)break;
}
for(int i=0;i<n;++i){
if(i)printf(" ");
printf("%d",num[i]);
}
return 0;
}