CF1305F. Kuroni and the Punishment

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给定一个序列，每次可以使$a_i+1,a_i-1$

求最小操作数使得$gcd(a_1,a_2…a_n)>1$

显然$gcd=2$,$ans<n$

假设有$k$个数,$+2,-2$,$2k<n,k<n/2$

所以可以知道一定有$n/2$个数$+1,-1,0$
随取选取一个数枚举他的质因数$n\log a_i$
如果实验次数为$m$,失败概率=$1-\frac{1}{2^m}$

代码 
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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>

#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10;
const int M = 70;
const int mod = 1e9 + 7;
int n;
ll a[N];
ll ans = N;
ll calc(ll x)
{
    ll res = 0;
    for (int i = 1; i <= n; i++)
    {
        if (a[i] < x)
            res += x - a[i] % x;
        else
            res += min(a[i] % x, x - a[i] % x);
        if (res > ans)
            break;
    }
    return res;
}
void solve(ll x)
{

    for (ll i = 2; i * i <= x; i++)
    {
        if (x % i == 0)
        {
            while (x % i == 0)
                x /= i;
            ans = min(calc(i), ans);
        }
    }
    if (x > 1)
        ans = min(calc(x), ans);
}
int main()
{
    scanf("%d", &n);
    srand(time(0));
    for (int i = 1; i <= n; i++)
        scanf("%lld", &a[i]);

    for (int i = 1; i <= M; i++)
    {
        ll x = a[1ll * rand() * rand() % n + 1];
        //cout << rand() % n + 1 << endl;
        solve(x + 1);
        solve(x);
        if (x > 1)
            solve(x - 1);
    }
    printf("%lld\n", ans);
}