题意:给一个时刻,求时针、分钟、秒针三者之间的夹角
思路:确定参照点,求出三者的绝对夹角,然后用差来得到它们之间的夹角,钝角情况用360。减去就行了。
#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];} const double PI = acos(-1.0); const int INF = 1e9 + 7; const double EPS = 1e-8; /* -------------------------------------------------------------------------------- */ int gcd(int a, int b) { return b? gcd(b, a % b) : a; } void pri(int a, int b) { int g = gcd(a, b); a /= g; b /= g; printf("%d", a); if (b > 1) printf("/%d", b); putchar(' '); } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int T, cas = 0; cin >> T; while (T --) { int h, m, s; scanf("%d:%d:%d", &h, &m, &s); h = h % 12; int H = 3600 * h + 60 * m + s, M = 720 * m + 12 * s, S = 720 * s; int tot = 360 * 120; int HM = abs(H - M), HS = abs(H - S), MS = abs(M - S); umin(HM, tot - HM); umin(HS, tot - HS); umin(MS, tot - MS); pri(HM, 120); pri(HS, 120); pri(MS, 120); puts(""); } return 0; }