【HDU】5778 - abs（思维）

来源：互联网发布：淘宝装修生成编辑：程序博客网时间：2024/06/10 01:04

点击打开题目

abs

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1359 Accepted Submission(s): 477

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

妈呀好紧张，bc结束之前5分钟AC了。

他要每个质因数个数都为2，那么开一下根号，就是说这个数质因数个数都为1，写一个函数就行了。

把最初的 n 开根号得 t ，如果 t * t == n 而且 t 不满足条件的话，那么分别从 t 的两边找，如果 t * t < n ，那么从 t + i 和 t - i + 1找（ i 从 0 开始）。

代码如下：（刚开始总是wa，小情况考虑不到，所以代码比较复杂，思路还是清楚的）

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <cmath>using namespace std;#define INF 0x3f3f3f3f#define CLR(a,b) memset(a,b,sizeof(a))bool check(__int64 x){    if (x == 1)        return false;    for (int i = 2 ; i <= sqrt(x) ; i++)    {        if (x % i == 0)        {            x /= i;            if (x % i == 0)                return false;        }    }    return true;}int main(){    int u;    __int64 n;    __int64 res;    scanf ("%d",&u);    while (u--)    {        scanf ("%I64d",&n);        if (n == 1)        {            printf ("3\n");            continue;        }        __int64 ans = sqrt(n);        if (check(ans) && ans * ans == n)        {            printf ("0\n");            continue;        }        if (ans * ans != n)        {            for (int i = 1 ; ; i++)            {                __int64 t1 = 0,t2 = 0;                if (check(ans+i))                {                    res = (ans+i)*(ans+i);                    res -= n;                    t1 = abs(res);                }                if (check(ans-i+1))                {                    res = (ans-i+1)*(ans-i+1);                    res -= n;                    t2 = abs(res);                }                if (t1 && t2)                {                    res = min (t1,t2);                    break;                }                else if (t1 || t2)                {                    res = max (t1,t2);                    break;                }            }        }        else        {            for (int i = 0 ; ; i++)            {                __int64 t1 = 0,t2 = 0;                if (check(ans+i))                {                    res = (ans+i)*(ans+i);                    res -= n;                    t1 = abs(res);                }                if (check(ans-i))                {                    res = (ans-i)*(ans-i);                    res -= n;                    t2 = abs(res);                }                if (t1 && t2)                {                    res = min (t1,t2);                    break;                }                else if (t1 || t2)                {                    res = max (t1,t2);                    break;                }            }        }        printf ("%I64d\n",abs(res));    }    return 0;}

0 0