• <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>
            隨筆 - 87  文章 - 279  trackbacks - 0
            <2006年9月>
            272829303112
            3456789
            10111213141516
            17181920212223
            24252627282930
            1234567

            潛心看書研究!

            常用鏈接

            留言簿(19)

            隨筆分類(81)

            文章分類(89)

            相冊

            ACM OJ

            My friends

            搜索

            •  

            積分與排名

            • 積分 - 216403
            • 排名 - 117

            最新評論

            閱讀排行榜

            評論排行榜

            Sailboat

            Problem H: Sailboat

            In the sailboat race, the contestant is requested to along with the prearrange path. Sailing ship's power comes from wind power and contestant's manpower. The wind power can completely used.

            In a competition, the contestants are requested to along with a 1/4 circles with radius R, the sailboat will goto east from south. During this process, the wind direction is straight from west to the east with constant speed and power.

            In order to maintain the travel direction, the athlete must adjust the sail to the vertical angle from movement direction in any time.

            If the speed of sailboat is proportional to the power at movement direction, the proportional factor is k. Supposes the wind power is f, the athlete manpower is h, please given the time of sailboat from the beginning to the end.

            Input

            The first line of each case consists of 4 double number, that is radius of path: R, wind power: f,athlete manpower: h and proportional factor:k. In order to avoid the floating point error, you needn't output the answer directly. The next line is a integer n, the following n lines gives a double value which is candidate answer.

            Output

            For each candidate of each case, Only "Yes" or "No" should be printed. Output "Yes" if the relative error to your answer is less than 3%, otherwise "No". For example, if the model answer is 100, and the candidate is 98 or 102, you should output "Yes". Output one blank line between neighboring case

            Sample Input

            1.0 2.0 1.0 1.0
            2
            0.35
            0.76
            

            Sample Output

            No
            Yes
            

            Problem Source: provided by skywind

            #include <iostream>
            #include 
            <cmath>
            using namespace std;

            const int MAXN = 100;
            const double PI = acos(-1.0);

            double R, F, H, K, ans;
            int n, cas;

            double func(double x) {
                
            return R / K / (H + F * cos(x));
            }


            double romberg(double a, double b, double EPS = 1e-6{
                
            double t[MAXN][MAXN] = {0}, tmp;
                
            int i, j, k, k2, m, m4;
                t[
            0][0= (func(a) + func(b)) * (b - a) / 2;
                k 
            = 1; k2 = 1;
                
            while (1{
                    tmp  
            = 0;
                    
            for (i = 1; i <= k2; i++{
                        tmp 
            += func(a + (2 * i - 1* (b - a) / (2 * k2));
                    }

                    t[
            0][k] = (t[0][k - 1+ tmp * (b - a) / k2) / 2;
                    
            for (m = 1, m4=4; m <= k; m++, m4 *= 4{
                        t[m][k 
            - m] = (m4 * t[m - 1][k - m + 1- t[m - 1][k - m]) / (m4 - 1);
                    }

                    
            if (fabs(t[k][0- t[k - 1][0]) < EPS) break;
                    k
            ++; k2 *= 2;
                }

                
            return t[k][0];
            }


            void solve() {
                
            double tmp;
                scanf(
            "%lf"&tmp);
                
            if (fabs(tmp - ans) / ans < 0.03) printf("Yes\n");
                
            else printf("No\n");
            }


            int main() {
                freopen(
            "2457.in""r", stdin);
                
            while (scanf("%lf%lf%lf%lf%d"&R, &F, &H, &K, &n) != EOF) {
                    
            if (cas) printf("\n");
                    
            else cas++;
                    ans 
            = romberg(0, PI/2);
                    
            while (n--{
                        solve();
                    }

                }

                
            return 0;
            }
            posted on 2007-10-20 01:02 閱讀(686) 評論(0)  編輯 收藏 引用 所屬分類: 算法&ACM
            精品久久久久国产免费| 国产精自产拍久久久久久蜜| 久久久久久久综合综合狠狠| 99久久人人爽亚洲精品美女| 四虎国产精品成人免费久久| 国产成人久久精品一区二区三区| 国内精品久久久久影院薰衣草 | 国产无套内射久久久国产| 久久久久综合国产欧美一区二区| 久久久亚洲裙底偷窥综合| 久久精品无码专区免费青青| 久久99亚洲综合精品首页| 亚洲色婷婷综合久久| 久久久久国产日韩精品网站| 色婷婷综合久久久久中文一区二区| 亚洲精品国产成人99久久| 久久久久高潮综合影院| 久久99国产一区二区三区| 99麻豆久久久国产精品免费| 亚洲精品成人网久久久久久| 久久精品国产秦先生| 久久婷婷五月综合97色 | 亚洲第一极品精品无码久久| 久久青青草原精品国产软件| 99久久无码一区人妻a黑| 欧美一区二区久久精品| 久久久久久狠狠丁香| 97久久国产露脸精品国产| 久久综合伊人77777麻豆| 草草久久久无码国产专区| 日韩精品久久久久久免费| 久久人妻AV中文字幕| 亚洲欧洲中文日韩久久AV乱码| 国产叼嘿久久精品久久| 精品九九久久国内精品| 国产精品99久久99久久久| 精品国产乱码久久久久久郑州公司 | 国内精品久久久久影院免费| 亚洲AV无码久久精品成人| 久久精品国产亚洲AV久| 综合久久久久久中文字幕亚洲国产国产综合一区首|