HDU 5778 ABS (brute Force enumeration)

Abs

Time limit:2000/1000 MS (java/others) Memory limit:131072/131072 K (java/others)
Total submission (s): 399 Accepted Submission (s): 140

Problem Descriptiongiven 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.

Inputthe first line of input was an integer T (1≤T≤ )
For each test case,the a single line contains, an integer x ( 1≤x≤ten)

Outputfor each testcase print the absolute value of y-x

Sample Input511124290871699579095

Sample Output23656724470 Description: Give you a long long range of number x, and then ask if you can find a number y, number y in the qualitative factor decomposition, each number just appeared two times, the output is abs (X-y).　　The puzzle: When the game did not come out, and then see the puzzle found so water = =. Official problem: As the Y-factor decomposition of each factorization appear 2 times, then Y is a complete square number, set Y=z*z, the title can be converted to Z, so that each factorization appear 1 times. We can brute Force enumeration z, check whether z conforms to requirements, obviously when Z is prime number is compliant, by the prime number theorem can get, Z's enumerator at Logn level complexity O (\sqrt[4]{n}log\sqrt[2]{n}?4??√?N???LOg ? 2 ?? √ ? n ? ?? ) < Span>< Span class= "Mord" > < Span class= "Baseline-fix" > just need to enumerate around X, the scope of the enumeration is very small ... Then z to determine whether the condition is satisfied is also very simple. Code:

1#include <cstdio>2#include <cmath>3#include <iostream>4#include <algorithm>5#include <vector>6#include <stack>7#include <cstring>8#include <queue>9#include <Set>Ten#include <string> One#include <map> A #defineINF 9223372036854775807 - #defineINF 9e7+5 - #definePI ACOs (-1) the using namespacestd; -typedefLong Longll; -typedefDoubledb; - Const intMAXN = 1e5 +5; + Const intMoD = 1e9 +7; - ConstDB EPS = 1e-9; +  A //judge that this number is not satisfied with the condition.  at BOOLOK (ll x) { -      for(LL i =2; I*i <= x; i++) { -         intnum =0; -          while(x% i = =0) {/*Here I meet while the condition must be prime, for what, and the prime number of the principle of the screening method is similar. */ -num++; -X/=i; in             if(Num >1)return false;//if the condition is not met about two times -         } to     } +     return true; - } the  * ll ABSs (ll x) { $     returnX >=0? X:-x;Panax Notoginseng } -  the voidsolve () { +ll N, ans; CIN >>N; All z = sqrt (n+0.5D); the     //cout << z << endl; +      for(LL i =0; ; i++) { -ll tmp = z +i; $         if(TMP * tmp >= n && OK (tmp)) {/*because it does not match n small, so to meet >= n This condition, the test sample when found 8 root is 2, and then 2 is to meet the conditions, at this time found the answer is not right. */ $Ans = ABSS (tmp*tmp-n); -             //cout << tmp << Endl; -              Break; the         } -     }Wuyi      for(LL i =0;; i++) { thell tmp = Z-i; -         if(OK (tmp)) { Wuans = min (ans, ABS (TMP*TMP-n)); -              Break; About         } $     } -     if(N <4) ans =4N/*less than 4 of the root is 1, and then the OK function to determine the 1 is satisfied with the condition, but 1 is not a prime number, so the special sentence. */ -cout << ans <<Endl; - } A  + intMain () the { -     //Cin.sync_with_stdio (false); $     //freopen ("isharp.in", "R", stdin); the     //freopen ("Hh.txt", "w", stdout); the     intT CIN >>T; the  the      while(t--) - solve (); in     return 0; the}

HDU 5778 ABS (brute Force enumeration)