HDU-5778 abs

最新推荐文章于 2022-10-16 04:57:01 发布

原创最新推荐文章于 2022-10-16 04:57:01 发布 · 326 阅读

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本文介绍了一种算法，用于寻找给定数值下满足特定条件的最接近平方数。该算法通过枚举逼近的方法，结合质因数判断，实现了对目标平方数的有效搜索，并提供了完整的C++实现代码。

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原题链接

abs

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1260 Accepted Submission(s): 439

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

Sample Output

设p1 = sqrt(x), p2 = p1 +1, P1向小方向枚举，p2向大方向枚举，找到一个p使p的每种质因子仅有一个且abs(p*p-x)达到最小.

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <map>
#include <cmath>
#define maxn 100005
#define INF 1000000000
using namespace std;
typedef long long ll;

int num[maxn], cnt;
ll v[maxn];
void prime(){
	
	for(ll i = 2; i < maxn; i++){
		if(num[i] == 0){
			v[cnt++] = i;
		}
		for(ll j = i * i; j < maxn; j += i)
		 num[j] = 1;
	}
}
bool judge(ll p){
	
	if(p < 2)
	  return false;
	for(int i = 0; i < cnt&& v[i] * v[i] <= p; i++){
		
		ll d = v[i] * v[i];
		if(p % d == 0)
		 return false;
	}
	return true;
}
void solve(ll x){
	
	ll p1 = sqrt(x);
	ll p2 = p1 + 1;
	while(1){
		
		if(p1 >= 2 && x - p1 * p1 < p2 * p2 - x){
			
			if(judge(p1)){
				printf("%I64d\n", x - p1 * p1);
			    return ;
			}
		    p1--;
		}
		else{
			if(judge(p2)){
				printf("%I64d\n", p2 * p2 - x);
		        return ;
			}
			p2++;
		}
	}
}
int main(){
	
//	freopen("in.txt", "r", stdin);
	int t;
	prime();
	scanf("%d", &t);
	while(t--){
		ll x;
		scanf("%I64d", &x);
		solve(x);
	} 
	return 0;
}