质因数的思维题

Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.

Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.

Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.

Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.

Input

The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.

Output

Output a single integer — the minimum possible X0.

Examples
Input
Copy
14
Output
6
Input
Copy
20
Output
15
Input
Copy
8192
Output
8191
Note

In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:

  • Alice picks prime 5 and announces X1 = 10
  • Bob picks prime 7 and announces X2 = 14.

In the second case, let X0 = 15.

  • Alice picks prime 2 and announces X1 = 16
  • Bob picks prime 5 and announces X2 = 20.

题目分析 : 给你一个X2, 让你去寻找一个X0, 并且要求 x0 是最小的,选取的规则是每次选取一个质因数,在变换到X的时候是让其扩大一定的倍数,使其刚好大于等于当前的x

思路分析 :对于我输入的一个X2,我们发现当它的质因数越大时,上一个 x1 可以选取的值就可以越小,因此预处理下每个数最大的质因数就可以了

代码示例 :

#define ll long long
const int maxn = 1e6+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

int f[maxn];

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int x;
    cin >> x;
    for(int i = 2; i <= 1000000; i++){
        if (!f[i]){
            for(int j = i+i; j <= 1000000; j += i){
                f[j] = i;
            }
        }
        f[i] = i - f[i] + 1;
    }
    //for(int i = 1; i <= x; i++) printf("%d ", f[i]);
    int ans = inf;
    //cout << f[x] << endl;
    for(int i = f[x]; i <= x; i++){
        ans = min(ans, f[i]);
    }
    cout << ans << endl;
    return 0;
}

 

posted @ 2018-03-12 17:31  楼主好菜啊  阅读(181)  评论(0编辑  收藏  举报