hdu 2225 The nearest fraction （数学题）

　　一道简单数学题，要求求出一个分母不超过m的最接近sqrt(n)的分数。

　　做法就是暴力枚举，注意中间过程不能用浮点数比较，误差要求比较高。

代码如下：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <cmath>
 6 
 7 using namespace std;
 8 
 9 typedef long long LL;
10 template<class T> T sqr(T x) { return x * x;}
11 template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a;}
12 
13 int main() {
14     LL n, m;
15     while (cin >> n >> m) {
16         LL bx = 1, by = 1;
17 //        long double r = sqrt((long double) n);
18 //        cout << r << endl;
19         for (LL i = 1; i <= m; i++) {
20             LL t = (int) sqrt((double) sqr(i) * n);
21             for (int d = 0; d < 2; d++) {
22 //                if (fabs((long double) (t + d) / i - r) < fabs((long double) bx / by - r)) {
23 //                    bx = t + d, by = i;
24 //                }
25                 if (abs(sqr(t + d) * sqr(by) - n * sqr(by) * sqr(i)) < abs(sqr(bx) * sqr(i) - n * sqr(by) * sqr(i))) bx = t + d, by = i;
26             }
27         }
28         LL GCD = gcd(bx, by);
29         cout << bx / GCD << "/" << by / GCD << endl;
30     }
31     return 0;
32 }

View Code

——written by Lyon

posted @ 2013-07-13 19:43 LyonLys 阅读(206) 评论(0) 收藏举报

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