• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>

            poj1905

            Expanding Rods
            Time Limit: 1000MS Memory Limit: 30000K
            Total Submissions: 8376 Accepted: 2058

            Description

            When a thin rod of length L is heated n degrees, it expands to a new length L'=(1+n*C)*L, where C is the coefficient of heat expansion.
            When a thin rod is mounted on two solid walls and then heated, it expands and takes the shape of a circular segment, the original rod being the chord of the segment.

            Your task is to compute the distance by which the center of the rod is displaced.

            Input

            The input contains multiple lines. Each line of input contains three non-negative numbers: the initial lenth of the rod in millimeters, the temperature change in degrees and the coefficient of heat expansion of the material. Input data guarantee that no rod expands by more than one half of its original length. The last line of input contains three negative numbers and it should not be processed.

            Output

            For each line of input, output one line with the displacement of the center of the rod in millimeters with 3 digits of precision.

            Sample Input

            1000 100 0.0001
                        15000 10 0.00006
                        10 0 0.001
                        -1 -1 -1
                        

            Sample Output

            61.329
                        225.020
                        0.000
                        

            Source

            推一下公式

            然后二分就可以了

            可以二分的有很多

            但是如果二分圓心角的話感覺特別簡單


            code

            #include <cstdio>
            #include 
            <cstdlib>
            #include 
            <cstring>
            #include 
            <cmath>
            #include 
            <ctime>
            #include 
            <cassert>
            #include 
            <iostream>
            #include 
            <sstream>
            #include 
            <fstream>
            #include 
            <map>
            #include 
            <set>
            #include 
            <vector>
            #include 
            <queue>
            #include 
            <algorithm>
            #include 
            <iomanip>
            using namespace std;
            double l,ll,n,c;
            int main()
            {
                
            double left,mid,right;
                
            while(scanf("%lf%lf%lf",&l,&n,&c)!=EOF)
                {
                    
            if(l==-1&&n==-1&&c==-1break;
                    
            if(l==0||n==0||c==0)
                    {
                        printf(
            "0.000\n");
                        
            continue;
                    }
                    ll
            =l*(1+n*c);
                    left
            =0;
                    right
            =acos(-1.0);
                    
            //二分角度
                    while(right-left>1e-12)
                    {
                        mid
            =(left+right)/2;
                        
            if(mid*l>2*ll*sin(mid/2))
                            right
            =mid;
                        
            else left=mid;
                    }
                    printf(
            "%.3lf\n",(1-cos(mid/2))*l/(2*sin(mid/2)));
                }
                
            return 0;
            }

            posted on 2012-08-02 17:02 jh818012 閱讀(204) 評論(0)  編輯 收藏 引用

            <2025年7月>
            293012345
            6789101112
            13141516171819
            20212223242526
            272829303112
            3456789

            導航

            統(tǒng)計

            常用鏈接

            留言簿

            文章檔案(85)

            搜索

            最新評論

            • 1.?re: poj1426
            • 我嚓,,輝哥,,居然搜到你的題解了
            • --season
            • 2.?re: poj3083
            • @王私江
              (8+i)&3 相當于是 取余3的意思 因為 3 的 二進制是 000011 和(8+i)
            • --游客
            • 3.?re: poj3414[未登錄]
            • @王私江
              0ms
            • --jh818012
            • 4.?re: poj3414
            • 200+行,跑了多少ms呢?我的130+行哦,你菜啦,哈哈。
            • --王私江
            • 5.?re: poj1426
            • 評論內容較長,點擊標題查看
            • --王私江
            久久久精品人妻无码专区不卡 | 99热精品久久只有精品| 亚洲国产成人久久精品影视| 久久久久亚洲AV成人网人人软件 | 欧美久久一区二区三区| 久久人人爽人人爽人人AV| 久久精品国产黑森林| 99久久久精品免费观看国产 | 久久香蕉国产线看观看精品yw| 久久精品嫩草影院| 久久香蕉超碰97国产精品| 欧洲性大片xxxxx久久久| 国产精品久久久久久久久久免费| 日韩精品久久无码中文字幕| 亚洲国产高清精品线久久| 一级做a爰片久久毛片人呢| 日韩精品久久久久久免费| 2021国内久久精品| 无码任你躁久久久久久久| 久久国产精品二国产精品| 久久国产精品国产自线拍免费| 久久婷婷成人综合色综合| 亚洲精品高清国产一线久久| 中文字幕无码av激情不卡久久| 久久国产成人午夜AV影院| 久久久精品久久久久特色影视| 国产日韩欧美久久| 久久国产香蕉视频| 人人狠狠综合88综合久久| 无码人妻少妇久久中文字幕| 久久久久一级精品亚洲国产成人综合AV区| 亚洲国产精品久久久久婷婷老年| 久久综合久久综合久久综合| 久久99精品综合国产首页| 9999国产精品欧美久久久久久 | 亚洲国产精品久久久久婷婷软件| 久久国产精品-久久精品| 久久综合综合久久狠狠狠97色88| 2021国产成人精品久久| 久久丝袜精品中文字幕| 99久久香蕉国产线看观香|