PE 443

题目大意：给定 g(4)=13 , g(n)=g(n−1)+gcd(n,g(n−1))(n>=4) ,求 g(1015)

打表发现大多数gcd的值都为1,考虑去找下一个gcd不为1的值
设当前位置 f(n)=g
即求 gcd(n+1+d,g+d)!=1
gcd(g−n−1,g+d)!=1
然后暴力枚举因数就行了

INF赋小了，调了好久QAQ

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<string>
#include<ctime>
#include<cmath>
#define LL long long
using namespace std;

LL n,g,aim = (LL)1e15,x,y,d,t,k;

LL gcd(LL a,LL b)
{
    return b ? gcd(b,a % b) : a;
}

void update(LL i)
{
    t = g / i * i;
    if (t < g) t += i;
    if (t < y) y = t;
}

int main()
{
    n = 4,g = 13;
    while (true)
    {
        x = g - n - 1;y = (LL)1e17;
        for (LL i = 2;i * i <= x;i ++)
            if (x % i == 0) update(i),update(x / i);
        update(x);
        d = y - g;
        if (n + d + 1 <= aim)
        {
            g += d;
            n += d + 1;
            g += gcd(g,n);
        }
        else
        {
            g += aim - n;
            break;
        }
    }
    cout << g << endl;

    return 0;
}