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
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 51 2 3
output
Copy
11100
input
Copy
4 210 3 5 3
output
Copy
32
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.
题解：如果直接暴力去找的话必然是会超时的，我们可以利用c++排序的功能排序好，然后进行比较每次输出最小的就大大提高了效率，从而不会是超时
代码：
#include#include #include #include using namespace std;int main(){ int n,m; cin>>n>>m; int a[100005]; for(int t=0;t