hdu 5778 abs（分解质因子，枚举平方数，数论）

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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

511124290871699579095

Sample Output

23656724470

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;}

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