【CUGBACM15级BC第六场 B】hdu 4982 Goffi and Squary Partition

Goffi and Squary Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1588    Accepted Submission(s): 527

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 because1 + 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 integersn and k (2≤n≤200000,2≤k≤30).

 

Output

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

 

Sample Input

2 24 222 4

 

Sample Output

NOYESYES

 

题目大意：给出n,k。问是否存在一个k个数字的序列，和为n，并且其中存在k-1个数字的和是一个完全平方数。

例如：{1, 5, 6, 10}, n为22，k为 3的时候， 1 + 5 + 6 + 10 = 22 and 1 + 5 + 10 = 16 = 4 × 4.所以符合

#include <vector>#include <map>#include <set>#include <algorithm>#include <iostream>#include <cstdio>#include <cmath>#include <cstdlib>#include <string>#include <cstring>using namespace std;int main(){    int n, k;    while (scanf("%d%d", &n, &k) != EOF)    {        int flag = 0;        int sum = k * (k - 1) / 2;        for (int i = 1; i * i < n; i++)        {            int squre = i * i;            int need = n - squre;            if (squre < sum)            {                continue;            }            if (need <= k - 1 && sum + k > n)            {                continue;            }            if (squre == sum + 1 && need == k)            {                continue;            }            flag = 1;            break;        }        if (flag)        {            printf("YES\n");        }        else        {            printf("NO\n");        }    }    return 0;}