HDU-4982-Goffi and Squary Partition【贪心】【构造】

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HDU-4982-Goffi and Squary Partition

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    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

题目链接：HDU-4982

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。

（理解了好久的题解才明白这么做，orz）

以下是代码：

#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++) //遍历可能的完全平方数         {            int squre = i * i;            int need = n - squre; //可能的第k个数            if (squre < sum) continue; //前 k-1个数的和大于完全平方数，则不符合             if (need <= k - 1 && sum + k > n) continue; //如果第k个数 <= k-1，那么构造这个完全平方数时用到的最小的数是k，并且此时总和>n，不符合            if (squre == sum + 1 && need == k) continue; //如果完全平方数 == sum+1，说明在构造完全平方数时需要用到k，而需要的第k个数也是k，产生矛盾            flag = 1;            break;        }        if (flag) printf("YES\n");        else printf("NO\n");    }    return 0;}