HDU-5778 abs

题目大意：

给定一个数x，求正整数y\geq 2y≥2，使得满足以下条件：

1.y-x的绝对值最小

2.y的质因数分解式中每个质因数均恰好出现2次。

解题思路：

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：

朴素写法：

#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;

const int MAXN = 1e5 + 10;

LL prime[MAXN];

void init()
{
memset(prime, 0, sizeof(prime));
for(int i = 2; i < MAXN; ++i){
if(!prime[i]) prime[++prime[0]] = i;
for(int j = 1; j <= prime[0] && prime[j] < MAXN / i; ++j){
prime[prime[j] * i] = 1;
if(i % prime[j] == 0) break;
}
}
}
bool isPrime(LL n){
for(int i = 2; i * i <= n; ++i){
if(n % i == 0) return 0;
}
return 1;
}
bool Judge(LL n){
for(int i = 1; i <= prime[0]; ++i){
if(n % prime[i] == 0){
n /= prime[i];
if(n % prime[i] == 0) return 0;
}
}
if(n == 1 || isPrime(n)) return 1;
else return 0;
}
LL solve(LL n){
LL d = sqrt(n + 0.5);
LL ans = 1e18;
for(LL i = d; i >= 2; --i){
if(Judge(i)){
ans = min(ans, n - (i * i));
break;
}
}
for(LL i = d + 1; ; ++i){
if(Judge(i)){
ans = min(ans, (i * i) - n);
break;
}
}
return ans;
}
int main(){
LL n, t;
cin >> t; init();
while(t--){
cin >> n;
cout << solve(n) << endl;
}
return 0;
}

Miller_Rabin + Pollard_rho写法：

#include <cmath>
#include <time.h>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 100 + 5;
const int S = 20;

int tot;
LL factor[maxn];

LL mod_mul(LL a, LL b, LL n){
LL res = 0;
while(b){
if(b & 1) res = (res + a) % n;
a = (a + a) % n;
b >>= 1;
}
return res;
}
LL mod_exp(LL a, LL b, LL n){
LL res = 1;
while(b){
if(b & 1) res = mod_mul(res, a, n);
a = mod_mul(a, a, n);
b >>= 1;
}
return res;
}
LL gcd(LL a, LL b){
if(a == 0) return 1;
if(a < 0) return gcd(-a, b);
while(b){
LL tmp = a % b;
a = b;
b = tmp;
}
return a;
}

LL pollard_rho(LL x, LL c){
LL i = 1, k = 2;
LL x0 = rand() % x;
LL y = x0;
while(true){
++i;
x0 = (mod_mul(x0, x0, x) + c) % x;
LL d = gcd(y - x0, x);
if(d != 1 && d != x) return d;
if(y == x0) return x;
if(i == k){
y = x0;
k += k;
}
}
}

bool miller_rabin(LL n) {
if(n == 2 || n == 3 || n == 5 || n == 7 || n == 11) return true;
if(n == 1 || !(n % 2) || !(n % 3) || !(n % 5) || !(n % 7) || !(n % 11)) return false;

LL x, pre, u = n - 1, k = 0;

while(!(u & 1)){
++k;
u >>= 1;
}

srand((LL)time(NULL));
for(int i = 0; i < S; ++i){                     //进行S次测试，S越大，结果越准确
x = rand() % (n - 2) + 2;                   //在[2, n)中取随机数
if(x % n == 0) continue;

x = mod_exp(x, u, n);                       //计算x^u % n
pre = x;
for(int j = 0; j < k; ++j){
x = mod_mul(x, x, n);
if(x == 1 && pre != 1 && pre != n - 1)
return false;
pre = x;
}
if(x != 1) return false;
}
return true;
}

void findfactor(LL n){
if(miller_rabin(n)){
factor[tot++] = n;
return;
}

LL p = n;
while(p >= n){
p = pollard_rho(p, rand() % (n - 1) + 1);
}
findfactor(p);
findfactor(n / p);
}

int main(){
LL t, x, tmp;
cin >> t;
while(t--){
cin >> x;
if(x == 1 || x == 2 || x == 3 || x == 4){
cout << (4 - x) << endl;
continue;
}

tmp = (LL)sqrt(x + 0.500);
LL ans = 1e18;
for(LL i = tmp; i > 1; --i){
tot = 0;
memset(factor, 0, sizeof(factor));
findfactor(i);
sort(factor, factor + tot);

int flag = 1;
for(int j = 1; j < tot; ++j){
if(factor[j] == factor[j-1]) {flag = 0; break;}
}
if(flag) {ans = min(ans, x - i * i); break;}
}
for(LL i = tmp + 1; ; ++i){
tot = 0;
memset(factor, 0, sizeof(factor));
findfactor(i);
sort(factor, factor + tot);

int flag = 1;
for(int j = 1; j < tot; ++j){
if(factor[j] == factor[j-1]) {flag = 0; break;}
}
if(flag) {ans = min(ans, i * i - x); break;}
}
cout << ans << endl;
}
return 0;
}