• <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
            久久伊人精品青青草原高清| 成人资源影音先锋久久资源网| 93精91精品国产综合久久香蕉| 72种姿势欧美久久久久大黄蕉| 夜夜亚洲天天久久| 国产日韩欧美久久| 亚洲国产欧美国产综合久久| 精品久久久久久久久午夜福利| 久久精品国产99久久久香蕉| 久久亚洲AV成人无码软件| 97久久超碰成人精品网站| 欧美伊人久久大香线蕉综合69| 久久精品亚洲精品国产色婷 | 久久久久久久久久久| 777米奇久久最新地址| 亚洲人成网站999久久久综合 | 国产高潮国产高潮久久久| 人妻中文久久久久| 久久综合久久综合久久| 亚洲精品无码久久久久去q | 超级碰碰碰碰97久久久久| 大香网伊人久久综合网2020| 囯产精品久久久久久久久蜜桃 | 久久国产免费观看精品3| 无夜精品久久久久久| 国产一久久香蕉国产线看观看| 国产精品久久久久久久久软件 | 亚洲国产精品婷婷久久| 2021少妇久久久久久久久久| 7777精品久久久大香线蕉| 伊人久久大香线蕉AV一区二区| 国产精品九九久久精品女同亚洲欧美日韩综合区 | 精品少妇人妻av无码久久| 亚洲AV无码久久精品成人| 国产欧美久久久精品影院| 欧美久久天天综合香蕉伊| 久久人妻少妇嫩草AV蜜桃| 亚洲国产综合久久天堂| 亚洲国产高清精品线久久| 久久中文字幕人妻熟av女| 99久久精品国产一区二区|