HDU 4982 Goffi and Squary Partition（BestCoder Round #6）

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:
[ol]

• 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.

[/ol]
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 2

4 2

22 4

 

Sample Output：

NO

YES

YES

 

题意：给出n和k的值，问能否找到一个序列满足以下条件：1.这个序列长度为k，这k个数的和是n；2.这k个数中存在任意k-1个数的和是任意一个数的平方。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<queue>
#include<algorithm>
using namespace std;

const int N=1e3+10;
const int M=50000;
const int INF=0x3f3f3f3f;

int main ()
{
    int n, k, sum, i, flag;
    int squre, remain;

    while (scanf("%d%d", &n, &k) != EOF)
    {
        flag = 0;
        sum = k*(k-1)/2; ///可以先令1~k-1这些数为前k-1个数（这是最小的k-1个数的和）

        for (i = 1; i*i < n; i++)
        {
            squre = i*i; ///完全平方数
            remain = n-squre; ///可能的第k个数

            if (sum > squre) continue; ///前k-1个数的和大于完全平方数，不符合题意
            if (remain <= k-1 && sum+k > n) continue; ///如果第k个数<=k-1，那么构造这个完全平方数时用到的最小的数是k，并且此时总和>n，不符合题意
            if (squre == sum+1 && remain == k) continue; ///如果完全平方数==sum+1，说明在构造完全平方数时需要用到k，而需要的第k个数也是k，产生矛盾

            flag = 1;
            break;
        }

        if (flag) printf("YES
");
        else printf("NO
");
    }

    return 0;
}