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 <bits/stdc++.h>
// 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<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
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<int> 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;
}