数论 gcd

看到这个题其实知道应该是和(a+k)(b+k)/gcd(a+k,b+k)有关，但是之后推了半天，思路全无。

然而。。有一个引理:

• gcd(a, b) = gcd(a, b - a) = gcd(b, b - a) (b > a)

证明一下：

令 gcd(a, b) = c, (b > a)

则有 a % c = 0, b % c = 0

那么 (a - b) % c = 0

令 gcd(a, b - a) = c', 假设c' != c

则有 a % c' = 0, (b - a) % c' = 0

则 (b % c' - a % c') % c' = 0, 所以 b % c' - a % c' = 0

所以 b % c' = 0

所以可以得出 c = c', 与假设矛盾, 则 c = c'.

同理可得 gcd(b, a - b) = c

证毕。

然后我们要求最小的k，那就枚举定值b-a的所有约数，看看a和b中小的那个数要凑成含这个约数的最小k是多少, 暴力找最大的lcm.

#include 
    
     //          9223372036854775807#define INF 2333333333333333333#define full(a, b) memset(a, b, sizeof a)using namespace std;typedef long long ll;inline int lowbit(int x){ return x & (-x); }inline int read(){    int X = 0, w = 0; char ch = 0;    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();    return w ? -X : X;}inline ll gcd(ll a, ll b){ return a % b ? gcd(b, a % b) : b; }inline ll lcm(ll a, ll b){ return a / gcd(a, b) * b; }template
     
      inline T max(T x, T y, T z){ return max(max(x, y), z); }template
      
       inline T min(T x, T y, T z){ return min(min(x, y), z); }template
       
        inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}int main(){    ll a, b, c;    cin >> a >> b;    if(a > b) swap(a, b);    c = b - a;    int n = (int)(sqrt(c) + 0.5);    vector
        
          v;    for(int i = 1; i <= n; i ++){        if(c % i == 0) v.push_back(i), v.push_back(c / i);    }    int k = 0; ll ans = INF;    for(int i = 0; i < v.size(); i ++){        int tmp = 0;        if(a % v[i] != 0) tmp = v[i] - a % v[i];        ll r = lcm(a + tmp, b + tmp);        if(r < ans) ans = r, k = tmp;    }    cout << k << endl;    return 0;}