[hdu4627 The Unsolvable Problem]数论

题意：给一个数n，找一个正整数x<n，使得x和n-x的最小公倍数最大。

思路：显然x和n-x越接近越好，gcd必须为1（贪心）。从大到小考虑x,如果n为奇数，则答案就是x=n/2，如果n为偶数，令n=2k，如果k为奇数，且大于1则x=k-2否则x=k，如果k为偶数则x=k-1。以上结论基于以下两个事实：

（1）相邻两个数的gcd为1

（2）相邻两个奇数的gcd为1

（3）1和1的gcd为1

写的时候没用到这些事实，直接从n/2向下枚举，当gcd为1时得到答案

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#include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; //#ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} //#endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} template<typename T> void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} template<typename T> void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} #if 0 const double PI = acos(-1.0); const int INF = 1e9 + 7; const double EPS = 1e-8; #endif /* -------------------------------------------------------------------------------- */ ll gcd(ll a, ll b) { return b? gcd(b, a % b) : a; } ll work(int n) { int x = n / 2; for (int i = x; i; i --) { if (gcd(i, n - i) == 1) return (ll)i * (n - i) / gcd(i, n - i); } } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int T, x; cin >> T; while (T --) { scanf("%d", &x); cout << work(x) << endl; } return 0; }