hdu 4859(思路题)

Posted 2020-08-01 AC菜鸟机

Goffi and Squary Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1171    Accepted Submission(s): 402

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

 

 

Source

BestCoder Round #6

 

构造数组出来，真的难想到，无比佩服能够想到的大牛。。分析都在代码里。。

/**
题意:给你k个数(各不相同的正整数),它们的和是n，问是否存在k-1个数的和是某个数的平方?
分析:我们假设其中k-1个数个数的平方为 m , m 应该不小于 k*(k-1)/2,不然就会有重复的 。
我们可以通过构造一个序列来判断是否满足条件。
*/
#include <stdio.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <vector>
using namespace std;
typedef long long LL;

int n,k;
bool judge(int m){
    int t = n-m, sum = (k-1)*k/2;
    if(sum>m) return false; ///k-1项最小都要 sum
    int count=0,x=0;
    for(int i=1;i<k-1;i++){ ///构造出一个序列包含剩下的序列中的 k - 2 项
        x++;
        if(t==x) x++;
        count+=x;
    }
    int lastnum = n-count-t; ///剩下的序列中的第 k-1 个数
    if(lastnum<=x) return false; ///如果最后一个数小于 x ，那么之前肯定存在过了
    x++;
    if(t==x||t==x+1){ ///如果 t == x 或者 t==x+1 那么当 第k-1个数等于t 的时候我们无法通过改变前k-2
                      ///项的值令 lastnum != t 了，如果t更大一些,我们是可以通过改变 k-1这个序列的值
                      ///来让 lastnum != t 的.
        if(lastnum==t) return false;
    }
    return true;
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF){
        bool flag = false;
        for(int i=1;i*i<n;i++){
            int m = i*i;
            if(judge(m)){
                flag = true;
                break;
            }
        }
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}