• [hdu5387 Clock]时钟夹角问题


    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;
    }
    
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  • 原文地址:https://www.cnblogs.com/jklongint/p/4728058.html
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