1088B Ehab and subtraction

B. Ehab and subtraction

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You're given an array aa. You should repeat the following operation kk times: find the minimum non-zero element in the array, print it, and then subtract it from all the non-zero elements of the array. If all the elements are 0s, just print 0.

Input

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The first line contains integers nn and kk (1≤n,k≤105)(1≤n,k≤105), the length of the array and the number of operations you should perform.

The second line contains nn space-separated integers a1,a2,…,ana1,a2,…,an (1≤ai≤109)(1≤ai≤109), the elements of the array.

Output

Print the minimum non-zero element before each operation in a new line.

Examples

input

Copy

3 5
1 2 3

output

Copy

input

Copy

4 2
10 3 5 3

output

Copy

3
2

Note

In the first sample:

In the first step: the array is [1,2,3][1,2,3], so the minimum non-zero element is 1.

In the second step: the array is [0,1,2][0,1,2], so the minimum non-zero element is 1.

In the third step: the array is [0,0,1][0,0,1], so the minimum non-zero element is 1.

In the fourth and fifth step: the array is [0,0,0][0,0,0], so we printed 0.

In the second sample:

In the first step: the array is [10,3,5,3][10,3,5,3], so the minimum non-zero element is 3.

In the second step: the array is [7,0,2,0][7,0,2,0], so the minimum non-zero element is 2.

题意：给了一个a数组，需要执行k个步骤，每次找出最小的那个非0数，让所有非0数减去这个最小数。如果所有数为0 ，就打印0.

题解：大体思路是对数组排个序，然后k次upper_bound(二分查找，找出第一个大于num的数的下标x），输出a[x]-num,如果a[x]不大于0，就跳出，输出剩余k个0,break后，如果while（k--）的话要特别注意k的值，k--会可能跳到负数，死循环，超时，这步菜鸡操作，昨晚把自己都写自闭了，才找到原因....

c++：

#include<bits/stdc++.h>
using namespace std;
int a[100010],n,k,num;;
int main()
{
    cin>>n>>k;
    for(int i=0; i<n; i++)
        cin>>a[i];
    sort(a,a+n);
    while(k--)
    {
        int x=upper_bound(a,a+n,num)-a;
        if(a[x]>0) cout<<a[x]-num<<endl;
        else
        {
             cout<<0<<endl;
             break;
        }
        num+=a[x]-num;
    }
    while(k--&&k>=0)
    {
        cout<<0<<endl;
    }
    return 0;
}

python：

n,k = map(int, input().split())
a = sorted(list(map(int, input().split())))
i,m = 0,0
while k:
    while a[i] - m <= 0 and i<n-1: i+=1
    print(a[i]-m)
    m+=a[i]-m
    k-=1

1088B Ehab and subtraction

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