You are given an integer x. Your task is to split the number x into exactly n strictly positive integers such that the difference between the largest and smallest integer among them is as minimal as possible. Can you?
The first line contains an integer T (1 ≤ T ≤ 100) specifying the number of test cases.
Each test case consists of a single line containing two integers x and n (1 ≤ x ≤ 109, 1 ≤ n ≤ 1000), as described in the statement above.
For each test case, print a single line containing n space-separated integers sorted in a non-decreasing order. If there is no answer, print - 1.
1 5 3
1 2 2
The strictly positive integers are the set defined as: . That is, all the integers that are strictly greater than zero: .
题目意思:对于T组输入输出,给你一个n将其分为k份,要求分成的份最大值和最小值相差最小。
解题思路:这道题很简单,我看了队友写的代码后发现这个题还挺有意思的,这里给出解释。
例如10分成4份,用10直接去整除4得到的是2,我们可以这样理解平均每一份是ans 2,即n/k得到的是一种平均的分配,但是这样还没有全部分完n,还会有余数m=n%k,因为m必然是比k小的数,也就可以理解为有m个分数需要在原来的ans上再加一个上1,这样得到的分配能够保证最大值和最小值相差最小。
1 #include2 #include 3 using namespace std; 4 int main() 5 { 6 int t,ans,m,i,k,n; 7 scanf("%d",&t); 8 while(t--) 9 {10 scanf("%d%d",&n,&k);11 if(n