HDU 5778 BestCoder Round #85 abs （暴力枚举）

abs

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 868    Accepted Submission(s): 309

Problem Description

Given a number x, ask positive integer y≥2, that satisfy the following conditions:
1. The absolute value of y - x is minimal
2. To prime factors decomposition of Y, every element factor appears two times exactly.

Input

The first line of input is an integer T ( 1≤T≤50)
For each test case，the single line contains, an integer x ( 1≤x≤1018)

Output

For each testcase print the absolute value of y - x

Sample Input

5 1112 4290 8716 9957 9095

Sample Output

23 65 67 244 70

Source

BestCoder Round #85

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题意：

给定一个数x，求正整数y，使得满足以下条件：
1.y-x的绝对值最小。
2.y的质因数分解式中每个质因数均恰好出现2次。

叫你求出y-x的最小绝对值。

官方题解：由于y质因数分解式中每个质因数均出现2次，那么y是一个完全平方数，设y=z*z，题目可转换成求z，使得每个质因数出现1次. 我们可以暴力枚举z，检查z是否符合要求，显然当z是质数是符合要求，由素数定理可以得，z的枚举量在logn级别复杂度

AC代码：

#include
using namespace std;
const long long INF = 0x7ffffffffll;
long long ans;
long long n;
bool solve(long long x) //判断这个数满不满足条件 
{
    if(x<2)
        return false;
    long long t=x;
    for(long long i=2;i*i<=x;i++) //暴力枚举质因数分解式 
    {
        if(x%i==0)
        {
            if(x%(i*i)==0) //出现了超过一次
            {
                return false;
            }
            x/=i;
        }
    }
    ans=min(ans,abs(t*t-n)); //最小绝对值 :y-x=(t^2-n) 
    return true;
}
int main ()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d",&n);
        long long x=(long long)(sqrt(n)+0.5);
        int flag=0;
        ans=INF;
        for(int i=0; ;i++)
        {
            if(solve(x+i))
                flag=1;
            if(solve(x-i))
                flag=1;
            if(flag==1)
                break;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}