标签：acm   数学   

题目：

Clock

4
00:00:00
06:00:00
12:54:55
04:40:00

0 0 0 
180 180 0 
1391/24 1379/24 1/2 
100 140 120 
Hint
每行输出数据末尾均应带有空格
 

题意：给一个时间，求出时针，分针，秒针两两之间的夹角。

思路：把时针，分针，秒针的角度都求出来即可，时针的角度等于(hh+mm/60+ss/3600)*30,分针的角度等于(mm+ss/60)*6,秒针的角度为6*ss,需要注意的是，数据用分数表示，所以两两求差的时候也应该是分数相减，然后求GCD使他们互质。当大于180°时用360°减去它。

代码：

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include<climits>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;

#define PB push_back
#define MP make_pair

#define REP(i,x,n) for(int i=x;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define FORD(i,h,l) for(int i=(h);i>=(l);--i)
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define OI(X) printf("%d",X);
#define RS(X) scanf("%s", (X))
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
#define Swap(a, b) (a ^= b, b ^= a, a ^= b)
#define Dpoint  strcut node{int x,y}
#define cmpd int cmp(const int &a,const int &b){return a>b;}

 /*#ifdef HOME
    freopen("in.txt","r",stdin);
    #endif*/
const int MOD = 1e9+7;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;
//#define HOME

int Scan()
{
	int res = 0, ch, flag = 0;

	if((ch = getchar()) == '-')				//判断正负
		flag = 1;

	else if(ch >= '0' && ch <= '9')			//得到完整的数
		res = ch - '0';
	while((ch = getchar()) >= '0' && ch <= '9' )
		res = res * 10 + ch - '0';

	return flag ? -res : res;
}
/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/


int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);

}

int main()
{int T;
RI(T);
char str[10];
while(T--)
{
    scanf("%s",str);
    int hh=0,mm=0,ss=0;
    hh=(str[0]-'0')*10+str[1]-'0';
    mm=(str[3]-'0')*10+str[4]-'0';
    ss=(str[6]-'0')*10+str[7]-'0';
    hh=hh%12;
    int h1=3600*hh-660*mm-11*ss;
    if(h1<0)
        h1=-h1;
    int h2=120;
    int g1=gcd(h1,h2);
    h1/=g1;
    h2/=g1;
    if(h1>h2*180)
        {h1=360*h2-h1;
    g1=gcd(h1,h2);
    h1/=g1;
    h2/=g1;}
    if(h2==1)
        printf("%d ",h1);
    else
        printf("%d/%d ",h1,h2);
    int s1=3600*hh+60*mm+ss-720*ss;
    if(s1<0)
        s1=-s1;
    int s2=120;
    int g3=gcd(s1,s2);
    s1/=g3;
    s2/=g3;
    if(s1>s2*180)
        {s1=360*s2-s1;
    g3=gcd(s1,s2);
    s1/=g3;
    s2/=g3;}
    if(s2==1)
        printf("%d ",s1);
    else
        printf("%d/%d ",s1,s2);
    int m1=60*mm-59*ss;
    int m2=10;
    if(m1<0)
        m1=-m1;
    int g2=gcd(m1,m2);
    m1/=g2;
    m2/=g2;
    if(m1>m2*180)
    {

    m1=360*m2-m1;
    g2=gcd(m1,m2);
    m1/=g2;
    m2/=g2;}
    if(m2==1)
    printf("%d \n",m1);
    else
        printf("%d/%d \n",m1,m2);
}



        return 0;
}

hdu5387 Clock(数学水题)

标签：acm   数学   

原文地址：http://blog.csdn.net/u013840081/article/details/48061025