HDU-5778 abs(数论)

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abs

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1163 Accepted Submission(s): 407

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

511124290871699579095

Sample Output

23656724470

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级别复杂度 O（ $\sqrt[4]{n}log\sqrt[2]{n}$ ）

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<limits.h>#include<math.h>#include<queue>#include<algorithm>using namespace std;#define maxn  100005__int64 a[maxn]={1,1};__int64 b[maxn];__int64  ans=-1;__int64  n,x;long long Abs(long long a){if(a<0)return -a;return a;}int main(){long long i,j,T;for(i=2;i<=100000;i++){if(a[i])    continue;for(j=i*i;j<=100000 && j>0;j+=i)     a[j]=1;}__int64 k=0;for(i=2;i<=100000;i++)      if(!a[i])        b[++k]=i;scanf("%I64d",&T);while(T--){long long tt;ans=-1;int flag;scanf("%I64d",&n);       x=(long long)sqrt(n);        for(i=x;i<=500+x;i++)        {        if(Abs(n-i*i)>ans && ans!=-1)        break;        tt=i;        for(j=1;j<=k;j++)        {        if(tt%b[j]==0)        {        tt/=b[j];        if(tt%b[j]==0)         break;}}if(ans==-1 && j==k+1)  ans=Abs(n-i*i);else if(j==k+1 && ans!=-1)  ans=min(ans,Abs(n-i*i));}for(i=x;i>=x-500;i--)        {        if(Abs(n-i*i)>ans && ans!=-1)        break;        tt=i;        for(j=1;j<=k;j++)        {        if(tt%b[j]==0)        {        tt/=b[j];        if(tt%b[j]==0)         break;}}if(ans==-1 && j==k+1)  ans=Abs(n-i*i);else if(j==k+1 && ans!=-1)  ans=min(ans,Abs(n-i*i));}if(n==1)       printf("3\n");    else if(n==2)      printf("2\n");    else    printf("%I64d\n",ans);    }   }

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