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最新推荐文章于 2025-08-09 19:02:03 发布

topc0o0der

最新推荐文章于 2025-08-09 19:02:03 发布

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分类专栏： HDU 文章标签： struct 360 ios

本文链接：https://blog.youkuaiyun.com/topc0o0der/article/details/5989365

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本文介绍了一种使用C++实现的高精度模拟时钟指针角度变化的方法，通过将一秒划分为10000份来确保计算精度达到10E-5级别。该方法考虑了时针、分针和秒针之间的角度变化，并通过模拟一天内每秒的变化来计算最大角度差。

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关键是精度；

3600*24=86400；

10E-4

题目要求的是10E-5

所以最少也要把一秒分为10份，但是我分到1000才对的。

为啥呢，其实算上四舍五入的话，题目要求E-6，而计算时应该更小，因为有累加效果。当然我的这个方法会超时，没事，打表就行了。

#include <iostream> #include <cstring> #include <stdio.h> #include <iomanip> using namespace std; int t; int min(int a,int b){ return a>b?b:a; } int abs(int a){ return a>0?a:-a; } //一秒钟时针转1个单位； //6/60/12=1/120->1 12 720 const int K=120*10000; const int day=24*60*60; const int cir=360*K; struct H{ int agle; H(){agle=0;} H operator ++(void){ ++agle; if(agle>=cir) agle-=cir; return *this; } }; struct M{ int agle; M(){agle=0;} M operator ++(){ agle+=12; if(agle>=cir) agle-=cir; return *this; } }; struct S{ int agle; S(){agle=0;} S operator ++(){ agle+=720; if(agle>=cir) agle-=cir; return *this; } }; int max_ag(H h,M m,S s){ int Min; Min=min(abs(h.agle-m.agle),abs(cir-abs(h.agle-m.agle))); Min=min(abs(h.agle-s.agle),Min); Min=min(abs(cir-abs(h.agle-s.agle)),Min); Min=min(abs(s.agle-m.agle),Min); Min=min(abs(cir-abs(s.agle-m.agle)),Min); return Min; } int Cnt[360]; void simulation(){ H h; M m; S s; for(int i=0; i<day; ++i){ for(int j=0; j<10000; ++j){ t=max_ag(++h,++m,++s)/K; ++Cnt[t]; } } } double ans[121]={100.000,98.340,96.694,95.061,93.443,91.839,90.249,88.672,87.110,85.562,84.027,82.507,81.001,79.508,78.030,76.565,75.114,73.676,72.252,70.841,69.445,68.063,66.694,65.340,64.000,62.673,61.361,60.062,58.778,57.507,56.251,55.008,53.779,52.562,51.360,50.172,48.997,47.838,46.692,45.560,44.443,43.339,42.249,41.173,40.111,39.064,38.030,37.010,36.002,35.008,34.029,33.063,32.111,31.174,30.250,29.340,28.444,27.563,26.695,25.841,25.001,24.175,23.363,22.565,21.779,21.007,20.250,19.506,18.776,18.060,17.359,16.671,15.997,15.338,14.693,14.061,13.444,12.841,12.252,11.675,11.112,10.563,10.028,9.507,9.000,8.507,8.028,7.562,7.111,6.674,6.251,5.842,5.446,5.065,4.697,4.342,4.001,3.674,3.361,3.062,2.777,2.506,2.249,2.005,1.776,1.561,1.360,1.172,0.999,0.840,0.695,0.562,0.444,0.340,0.249,0.173,0.110,0.062,0.027,0.007,0.000}; int main(){ //simulation(); //for(int i=121; i>-1; --i) //Cnt[i]+=Cnt[i+1]; //freopen("ans.txt","w",stdout); //cout<<Cnt[n]<<endl; // for(int i=0; i<121; ++i) // printf("%.3lf,",(double)Cnt[i]*100/(day*10000)); int n; while(cin>>n,n!=-1){ cout<<setiosflags(ios::fixed)<<setprecision(3)<<ans[n]<<endl; } //fflush(stdout); }

这个是比较好的代码：

http://www.acmdiy.net/Code/u/kalxd/HDU/1006