大致题意:每次操作可以使得序列某个数-1/+1,问最少多少次操作使得全局gcd>1
死也想不到随机化。。
我们考虑到 (gcd == 2) 的情况,显然答案的上界是 (n)。
我们考虑答案是不可能超过 (n) 的,那么至少 (frac{n}{2}) 是 0 或 1 。
然后直接随机 (x) 和 (x - 1) , (x + 1) 就可以了。
每次选中的概率是 (frac{1}{2})
所以你选 (m) 个数,那么正确率就是 (1 - frac{1}{2^m})
// powered by c++11
// by Isaunoya
#include <bits/stdc++.h>
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
#define Tp template
using pii = pair<int, int>;
#define fir first
#define sec second
Tp<class T> void cmax(T& x, const T& y) {if (x < y) x = y;} Tp<class T> void cmin(T& x, const T& y) {if (x > y) x = y;}
#define all(v) v.begin(), v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
Tp<class T> void sort(vector<T>& v) { sort(all(v)); } Tp<class T> void reverse(vector<T>& v) { reverse(all(v)); }
Tp<class T> void unique(vector<T>& v) { sort(all(v)), v.erase(unique(all(v)), v.end()); }
const int SZ = 1 << 23 | 233;
struct FILEIN { char qwq[SZ], *S = qwq, *T = qwq, ch;
#ifdef __WIN64
#define GETC getchar
#else
char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
#endif
FILEIN& operator>>(char& c) {while (isspace(c = GETC()));return *this;}
FILEIN& operator>>(string& s) {while (isspace(ch = GETC())); s = ch;while (!isspace(ch = GETC())) s += ch;return *this;}
Tp<class T> void read(T& x) { bool sign = 0;while ((ch = GETC()) < 48) sign ^= (ch == 45); x = (ch ^ 48);
while ((ch = GETC()) > 47) x = (x << 1) + (x << 3) + (ch ^ 48); x = sign ? -x : x;
}FILEIN& operator>>(int& x) { return read(x), *this; } FILEIN& operator>>(ll& x) { return read(x), *this; }
} in;
struct FILEOUT {const static int LIMIT = 1 << 22 ;char quq[SZ], ST[233];int sz, O;
~FILEOUT() { flush() ; }void flush() {fwrite(quq, 1, O, stdout); fflush(stdout);O = 0;}
FILEOUT& operator<<(char c) {return quq[O++] = c, *this;}
FILEOUT& operator<<(string str) {if (O > LIMIT) flush();for (char c : str) quq[O++] = c;return *this;}
Tp<class T> void write(T x) {if (O > LIMIT) flush();if (x < 0) {quq[O++] = 45;x = -x;}
do {ST[++sz] = x % 10 ^ 48;x /= 10;} while (x);while (sz) quq[O++] = ST[sz--];
}FILEOUT& operator<<(int x) { return write(x), *this; } FILEOUT& operator<<(ll x) { return write(x), *this; }
} out;
#define int long long
set < int > s ;
signed main() {
// code begin.
auto get = [&](int x) {
for(int i = 2 ; i <= sqrt(x) ; i ++) {
if(! (x % i)) {
s.insert(i) ;
while(! (x % i)) x /= i ;
}
}
if(x > 1) s.insert(x) ;
} ;
int n ;
in >> n ;
vector < int > a(n , 0) ;
for(int i = 0 ; i < n ; i ++) in >> a[i] ;
sort(a.begin() , a.end()) ;
int m = min(n , 30ll) ;
random_shuffle(a.begin() , a.end()) ;
for(int i = 0 ; i < m ; i ++) get(a[i]) ;
random_shuffle(a.begin() , a.end()) ;
for(int i = 0 ; i < m ; i ++) get(a[i] + 1) ;
random_shuffle(a.begin() , a.end()) ;
for(int i = 0 ; i < m ; i ++) get(a[i] - 1) ;
auto solve = [&](int x) {
int cnt = 0 ;
for(int i = 0 ; i < n ; i ++) {
if(a[i] < x)
cnt += x - a[i] ;
else
cnt += min(a[i] % x , x - a[i] % x) ;
}
return cnt ;
} ;
int ans = n ;
while(s.size()) {
cmin(ans , solve(* s.begin())) ;
s.erase(s.begin()) ;
}
out << ans << '
' ;
return 0;
// code end.
}