HDU 4982 Goffi and Squary Partition

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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): 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:
  • 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 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);
      // 1 k-1
      // 1 k-1 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
    "); else printf("NO
    "); } return 0; }