hdu 5778 abs（分解质因子，枚举平方数，数论）

abs

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 157    Accepted Submission(s): 63


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
题意：给你x求一个y满足abs(y-x)最小并且y分解质因子后所有的指数都是2，输出abs(y-x)

思路：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

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int devide(long long n)
{
int cnt=0;
for(long long i=2; i*i<=n; i++)
{
if(n%i==0)
{
int num=0;
while(n%i==0)
{
num++;
n/=i;
if(num>1) return 0;
}
if(num!=1) return 0;
}
}
return 1;
}
long long solve(long long n)
{
if(n<=4) return 4-n;
long long t=sqrt(n),m,l,ans;
for(l=0;; l++)
{
m=t+l;
if(m*m>=n&&devide(m))
{
ans=abs(m*m-n);
break;
}
}
long long k=n-ans;
for(long long i=0;;i++)
{
m=t-i;
if(m*m<=k) break;
if(devide(m)) ans=min(ans,abs(m*m-n));
}
return ans;
}
int main()
{
int T;
long long n;
scanf("%d",&T);
while(T--)
{
scanf("%I64d",&n);
printf("%I64d\n",solve(n));
}
return 0;
}