Goffi and Squary Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 856    Accepted Submission(s): 50

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:

1. the sum of k positive integers is equal to n

1. 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 2 4 2 22 4   Sample Output NO YES YES  

#include <vector> 
#include <list> 
#include <map> 
#include <set> 
#include <deque> 
#include <queue> 
#include <stack> 
#include <bitset> 
#include <algorithm> 
#include <functional> 
#include <numeric> 
#include <utility> 
#include <sstream> 
#include <iostream> 
#include <iomanip> 
#include <cstdio> 
#include <cmath> 
#include <cstdlib> 
#include <cctype> 
#include <string> 
#include <cstring> 
#include <ctime> 

using namespace std;

#define _int64 long long

int main()
{
  int n,k,tmp,sum,rem,b1,i;
  while (scanf("%d%d",&n,&k)!=EOF)
  {
    sum=(k*(k-1)/2);
　　//假设这个序列是从1到k-1的
　　//此处记录1到k-1的和
    b1=0;
　　//flag
    for (i=1;i*i<n;i++)
    {
      rem=n-i*i;
      tmp=i*i;
      if (tmp<sum) continue;
      if ((rem<=k-1)&&(tmp-sum<k-rem)) continue;
      //if ((tmp==sum)&&(rem<=k-1)) continue;
      if ((tmp==sum+1)&&(rem==k)) continue;
      b1=1;
      //cout<<i<<endl;
      break;
    }
    if (b1==1) printf("YES\n");
    else printf("NO\n");
  }
  return 0;
}