HDU 1006 Tick and Tick 模拟

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题意：就是问24小时中秒针，分针，时针的夹角两两大于d的时刻占总时刻的比重。

思路：模拟。我是求出秒针，分针，时针两两之间的速度，然后每次累加一个根据这个速度可以转180。的区间，然后每次替换掉一个结束时刻最早的区间，直到有一个区间的开始时刻大于24∗60∗60s。

坑点：时间不是连续的，不是1s,1s的走的，而是不停的在走，所以就不能计算出每一秒的相隔角度，再计算。

http://acm.hdu.edu.cn/showproblem.php?pid=1006

/*********************************************    Problem : HDU 1006    Author  : NMfloat    InkTime (c) NM . All Rights Reserved .********************************************/#include <map>#include <set>#include <queue>#include <cmath>#include <ctime>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#define rep(i,a,b)  for(int i = a ; i <= b ; i ++)#define rrep(i,a,b) for(int i = b ; i >= a ; i --)#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)#define cls(a,x)   memset(a,x,sizeof(a))#define eps 1e-8using namespace std;const int MOD = 1e9+7;const int INF = 0x3f3f3f3f;const int MAXN = 1e5;const int MAXE = 2e5;typedef long long LL;typedef unsigned long long ULL;int T,n,m,k;double degree;void input() {}double max_d(double a1,double a2,double a3) {    double max_deg = a1;    if(a2 > max_deg) max_deg = a2;    if(a3 > max_deg) max_deg = a3;    return max_deg;}double min_d(double a1,double a2,double a3) {    double min_deg = a1;    if(a2 < min_deg) min_deg = a2;    if(a3 < min_deg) min_deg = a3;    return min_deg;}void solve() {    double v12 = 6 - 1.0/10;    double v13 = 6 - 1.0/120;    double v23 = 1.0/10 - 1.0/120;    double base1,base2,base3,t1,t2,t3,t12,t13,t23,base12,base13,base23;    t1 = base1 = degree / v12 ; t2 = base2 = degree / v13; t3 = base3 = degree / v23;    t12 = 180 / v12; t13 = 180 / v13; t23 = 180 / v23;    base12 = t12 - t1 ; base13 = t13 - t2; base23 = t23 - t3;    double limit = 60 * 60 * 24;    int cnt1 = 1 , cnt2 = 1, cnt3 = 1;    double cro = 0;    double ans = 0;    while(1) {        //区间t1-t12 t2-t13 t3-t23        if(t1 > limit || t2 > limit || t3 > limit) break;        cro = min_d(t12,t13,t23) - max_d(t1,t2,t3);        if(cro < 0) cro = 0; ans += cro;        if(t12 <= t13 && t12 <= t23) {            if(cnt1 % 2) { t1 = t12 ; t12 += base12 ; }            else { t1 += 180 / v12 + base1; t12 = t1 + base12 ; }            cnt1 ++;        }        else if(t13 <= t12 && t13 <= t23) {            if(cnt2 % 2) { t2 = t13 ; t13 += base13 ; }            else { t2 += 180 / v13 + base2; t13 = t2 + base13 ; }            cnt2 ++;        }        else {            if(cnt3 % 2) { t3 = t23;  t23 += base23 ; }            else { t3 += 180 / v23 + base3; t23 = t3 + base23 ; }            cnt3 ++;        }    }    //printf("%d %d %d\n",cnt1,cnt2,cnt3);    printf("%.3lf\n",ans/limit*100);}int main(void) {    //freopen("a.in","r",stdin);    //scanf("%d",&T);    //while(T--) {    //while(~scanf("%d %d",&n,&m)) {    while(scanf("%lf",&degree),degree+1) {        input();        solve();    }    return 0;}

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