题目链接：http://codeforces.com/problemset/problem/1152/C
题目大意：
给定自然数 a，b，求最小自然数 k，使得 lcm(a + k, b + k) 最小。
分析：
根据《九章算术》“更相减损法”：
$$
gcd(a, b) = gcd(a, a - b)，a > b
$$
于是有：
$$
egin{align*}
lcm(a+k, b+k) &= frac{(a + k) * (b + k)}{gcd(a + k, b + k)} \
&= frac{(a + k) * (b + k)}{gcd(a + k, a - b)}
end{align*}
$$
又 gcd(a + k, a - b) 的值必定是 a - b 的某一个因子，所以只要枚举 a - b 的所有因子即可。
对于每一个因子 d，并不需要去求使得 gcd(a + k, a - b) == d 的最小 k，而只需要求使得 d|(a + k) 的最小 k 即可，因为如果此时的 gcd(a + k, a - b) > d，也就是说还有其他因子，那么这种情况在后面肯定会枚举到，并且会覆盖此时的答案。
代码如下：
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0); 5 #define Rep(i,n) for (int i = 0; i < (n); ++i) 6 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 12 13 #define pr(x) cout << #x << " = " << x << " " 14 #define prln(x) cout << #x << " = " << x << endl 15 16 #define LOWBIT(x) ((x)&(-x)) 17 18 #define ALL(x) x.begin(),x.end() 19 #define INS(x) inserter(x,x.begin()) 20 21 #define ms0(a) memset(a,0,sizeof(a)) 22 #define msI(a) memset(a,inf,sizeof(a)) 23 #define msM(a) memset(a,-1,sizeof(a)) 24 25 #define MP make_pair 26 #define PB push_back 27 #define ft first 28 #define sd second 29 30 template<typename T1, typename T2> 31 istream &operator>>(istream &in, pair<T1, T2> &p) { 32 in >> p.first >> p.second; 33 return in; 34 } 35 36 template<typename T> 37 istream &operator>>(istream &in, vector<T> &v) { 38 for (auto &x: v) 39 in >> x; 40 return in; 41 } 42 43 template<typename T1, typename T2> 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 45 out << "[" << p.first << ", " << p.second << "]" << " "; 46 return out; 47 } 48 49 inline int gc(){ 50 static const int BUF = 1e7; 51 static char buf[BUF], *bg = buf + BUF, *ed = bg; 52 53 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 54 return *bg++; 55 } 56 57 inline int ri(){ 58 int x = 0, f = 1, c = gc(); 59 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 60 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 61 return x*f; 62 } 63 64 typedef long long LL; 65 typedef unsigned long long uLL; 66 typedef pair< double, double > PDD; 67 typedef pair< int, int > PII; 68 typedef pair< string, int > PSI; 69 typedef set< int > SI; 70 typedef vector< int > VI; 71 typedef map< int, int > MII; 72 typedef pair< LL, LL > PLL; 73 typedef vector< LL > VL; 74 typedef vector< VL > VVL; 75 const double EPS = 1e-10; 76 const LL inf = 0x7fffffff; 77 const LL infLL = 0x7fffffffffffffffLL; 78 const LL mod = 1e9 + 7; 79 const int maxN = 5e5 + 7; 80 const LL ONE = 1; 81 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 82 const LL oddBits = 0x5555555555555555; 83 84 inline LL gcd(LL x,LL y){ 85 LL t; 86 if(!(x && y))return -1; 87 while(y){ 88 t=x%y; 89 x=y; 90 y=t; 91 } 92 return x; 93 } 94 95 inline LL lcm(LL x,LL y){ 96 LL t = gcd(x,y); 97 if(t == -1)return -1; 98 return x/t*y; 99 } 100 101 LL a, b, minLcm = infLL, k, ans; 102 103 VL factors; 104 inline void getFactors(int x) { 105 for(int i = 1; i * i <= x; ++i) { 106 if(x % i == 0) { 107 factors.push_back(i); 108 if(i != x / i)factors.push_back(x / i); 109 } 110 } 111 } 112 113 inline LL getMinK(LL d) { 114 return (d - (a % d)) % d; 115 } 116 117 int main(){ 118 INIT(); 119 cin >> a >> b; 120 if(a < b) swap(a, b); 121 getFactors(a - b); 122 123 Rep(i, factors.size()) { 124 k = getMinK(factors[i]); 125 LL tmp = (a + k) * (b + k) / factors[i]; 126 if(minLcm > tmp) { 127 minLcm = tmp; 128 ans = k; 129 } 130 } 131 cout << ans << endl; 132 return 0; 133 }