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): 1442    Accepted Submission(s): 479


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：输入整数n和k，要求把n分成k个数之和的形式，其中存在k-1个数之和为一个完全平方数，而且这k个数各不相同。
分析：我们尝试枚举那个完全平方数 S,然后看能否将他拆分为 K-1 个数,并且不用到N-S 这一步可以用贪心+一次调整来搞定。为了保证 K-1 个数都不同,我们尝试尽量

用 1,2,3...这些连续自然数来构造,如果 N-S 出现在这些数中,那么将 N-S 移除,再新加一个数。最后一个数由S-sum（1~k-2）(包括调整过的)来得到。

1.如果sum值大于S值，可以分成两种情况来看

1.1 前k-2个数中不存在N-S，那么原数列为1，2，3，....，k-2，其中的和大于等于S值，且最小的数为1，没有剩余的空间减少这k-2个数的和

1.2 前k-2个数中存在N-S，设x等于N-S那么原数列为1，2，....x-1，x+1，.....，k-1，其中多出来的空间为避免N-S，同样不存在剩余空间减少和

2.如果倒数最后一个数在前面k-2个数中出现，由上面结论可知，必定存在冲突，且无法调整
3.如果倒数最后一个数与N-S相等，那么可以使得倒数第一个数-1和倒数第二个数+1，这样的调整代价是最小的，如果这样的处理方式仍存在冲突，就为错

#include <cstdio>
using namespace std;
int pnt[35],top;
int n,k;
bool check(int x)
{
    int sum=0,top=0;
    int r=n-x,cc=0,cnt=1;
    pnt[top++]=0;
    for(int i=0; i<k-2; i++)
    {
        cc++;
        if(cc==r) cc++;
        pnt[top++]=cc;
        sum+=cc;
    }
    if(sum>=x) return false;
      pnt[top]=x-sum;
    if(pnt[top]<=pnt[top-1]) return false;
    if(pnt[top]==r)
    {
        if(pnt[top-1]+1>=pnt[top]-1) return false;
    }
    return true;
}
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        int flag=0;
        for(int i=1; i<=2000; i++)
        {
            if(i*i>=n) break;
            if(check(i*i))
            {
                printf("%d
",i*i);
                printf("YES
");
                flag=1;
                break;
            }
        }
        if(flag) continue;
        printf("NO
");
    }
    return 0;
}

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