BestCoder Round #6（1002）hdu4982（贪心）

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Goffi and Squary Partition

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18 Accepted Submission(s): 12

Problem Description

Recently, Goffi is interested in squary partition of integers.

A set X of k distinct positive integers is called squary partition of n if and only if it satisfies the following conditions:

the sum of k positive integers is equal to n
one of the subsets of X containing k−1 numbers sums up to a square of integer.

For example, a set {1, 5, 6, 10} is a squary partition of 22 because 1 + 5 + 6 + 10 = 22 and 1 + 5 + 10 = 16 = 4 × 4.

Goffi wants to know, for some integers n and k, whether there exists a squary partition of n to k distinct positive integers.

Input

Input contains multiple test cases (less than 10000). For each test case, there's one line containing two integers n and k (2≤n≤200000,2≤k≤30).

Output

For each case, if there exists a squary partition of n to k distinct positive integers, output "YES" in a line. Otherwise, output "NO".

Sample Input

2 24 222 4

Sample Output

NOYESYES

题意：构造k个数，使得总和为n，且存在k-1个数的总和为一个完全平方数，能构造输出YES，否则输出NO

思路：贪心，先找到小于n且最接近n的完全平方数x，然后构造前k-1个数使得总和为x，那么第k个数固定为y=n-x

做到这里就可以分情况讨论了

1. y不等于1~k-1里的数，如果 y 不等于 k 则有解，如果 y=k ，如果x-k*(k-1)/2=1，则无解，否则有解

2. y等于1~k-1里的数，显然y不可能等于k了，则将1~k-1里等于y的数替换为k即可，只要前k-1个数的总和<=x则有解，否则无解

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