hznu 1464: ST Task（算角度，注意精度）

最新推荐文章于 2021-02-25 17:11:37 发布

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本文深入探讨了AI音视频处理领域中的关键技术，特别是视频分割与语义识别。通过详细解释这些技术的工作原理、应用案例及实际效果，旨在为读者提供全面的理解和洞察。

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You get a ST task, that is: given a stick one end of who is moored on the ground, you are asked to turn over the stick by holding the other end. When it reaches the ground again, the task is finished. It is true that on the process, the stick is always on the same plane vertical the ground. And on this plane, there is light from up to down, so that we can see on the ground a line of shadow. Look at the picture:

In order to express the shadow part and the un-shadow (light space) part, to simple the problem we just need to express the length that 2 times of the length of the stick where the shadow may occur.

Now, give the problem: the stick on the beginning is on the left of the moored point, and we turn it on certain angular speed, using a ‘S’ to denote one unit of the light space and a ‘T’ for one unit of the shadow line. Besides that, a real number is needed to tell the scale between the shadow line and the full line where shadow may be.

输入

There is only one case. Two integers L (0 < L ≤ 25) and V (0 < V ≤ 90) is given.

L is the length of the stick; V is the angular speed of the turning task, in angle per second.

输出

For every second during the task, you are asked to tell the shape of the shadow on the ground. See the sample: ‘S’ for the light space and ‘T’ for the shadow.

样例输入

25 15

样例输出

TTTTTTTTTTTTTTTTTTTTTTTTTSSSSSSSSSSSSSSSSSSSSSSSSS  0.50000
STTTTTTTTTTTTTTTTTTTTTTTTSSSSSSSSSSSSSSSSSSSSSSSSS  0.48296
SSSTTTTTTTTTTTTTTTTTTTTTTSSSSSSSSSSSSSSSSSSSSSSSSS  0.43301
SSSSSSSTTTTTTTTTTTTTTTTTTSSSSSSSSSSSSSSSSSSSSSSSSS  0.35355
SSSSSSSSSSSSTTTTTTTTTTTTTSSSSSSSSSSSSSSSSSSSSSSSSS  0.25000
SSSSSSSSSSSSSSSSSSSTTTTTTSSSSSSSSSSSSSSSSSSSSSSSSS  0.12941
SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS  0.00000
SSSSSSSSSSSSSSSSSSSSSSSSSTTTTTTSSSSSSSSSSSSSSSSSSS  0.12941
SSSSSSSSSSSSSSSSSSSSSSSSSTTTTTTTTTTTTTSSSSSSSSSSSS  0.25000
SSSSSSSSSSSSSSSSSSSSSSSSSTTTTTTTTTTTTTTTTTTSSSSSSS  0.35355
SSSSSSSSSSSSSSSSSSSSSSSSSTTTTTTTTTTTTTTTTTTTTTTSSS  0.43301
SSSSSSSSSSSSSSSSSSSSSSSSSTTTTTTTTTTTTTTTTTTTTTTTTS  0.48296
SSSSSSSSSSSSSSSSSSSSSSSSSTTTTTTTTTTTTTTTTTTTTTTTTT  0.50000

http://hsacm.cn/JudgeOnline/problem.php?id=1464

#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<vector>
#include<cmath>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#include<queue>
#include<set>
#define ll long long
#define eps 1e-7    //本题注意精度问题 
using namespace std;
int main(){
    int n,m;
    cin>>n>>m; //n是长度，m是角速度 
    int x[1001];
    int p=-1;
    for(int i=0;i*m<=180;++i){
        double u=n*cos(3.1415926535*(double)i*m/180); //注意cos里面放的不是角度而是角度/180*π 
        int w=u+eps; //取整 
        if(u>=0){
	        if(u-w+eps>=0.5) //四舍五入 
	            w++;
	        int s=n-w;
	        for(int i=0;i<s;++i)
	            cout<<"S";
	        for(int i=0;i<w;++i)
	            cout<<"T";
	        for(int i=0;i<n;++i)
	            cout<<"S";
	        cout<<"  ";
	        printf("%.5lf\n",(double)u/(n+n)+eps); //T的比重 
		}
        else{
	        for(int i=0;i<n;++i)
	            cout<<"S";
	        u*=-1;
	        int w=u+eps;
	        if(u-w+eps>=0.5)
	            w++;
	        for(int i=0;i<w;++i)
	            cout<<"T";
	        int s=n-w;
	        while(s--)
	            cout<<"S";
	        cout<<"  ";
	        printf("%.5lf\n",(double)u/(n+n)+eps);
		}
	}
    return 0;
}