B. Ehab and subtraction
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
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
3 5 1 2 3
output
1 1 1 0 0
input
4 2 10 3 5 3
output
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：大概就是说给个非0数组，找其中的非0最小值，然后让所有数组元素剪去这个最小值，如此反复k次，如果所有元素都变成了0， 那就直接输出0就好了。排个序，把每次要减的值累加起来就好了。
#include<iostream>
#include<memory.h>
#include<algorithm>
using namespace std;
const int maxn = 1e5+5;
int n, k;
int a[maxn];
int main()
{
scanf("%d%d", &n, &k);
for(int i = 0;i < n;i ++)
scanf("%d", &a[i]);
sort(a, a + n);
int cnt = 0, top = 0;
while(k -- && top < n)
{
int t = a[top];
printf("%d\n", t - cnt);
while(t == a[top]) top ++;
cnt += t - cnt;
}
k ++;
while(k --)
printf("0\n");
return 0;
}