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
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
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
5 1112 4290 8716 9957 9095
Sample Output
23 65 67 244 70
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;
}
