hdu 2225 The nearest fraction （数学题）

Problem - 2225

　　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：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>

using namespace std;

typedef long long LL;
template<class T> T sqr(T x) { return x * x;}
template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a;}

int main() {
    LL n, m;
    while (cin >> n >> m) {
        LL bx = 1, by = 1;
//        long double r = sqrt((long double) n);
//        cout << r << endl;
        for (LL i = 1; i <= m; i++) {
            LL t = (int) sqrt((double) sqr(i) * n);
            for (int d = 0; d < 2; d++) {
//                if (fabs((long double) (t + d) / i - r) < fabs((long double) bx / by - r)) {
//                    bx = t + d, by = i;
//                }
                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;
            }
        }
        LL GCD = gcd(bx, by);
        cout << bx / GCD << "/" << by / GCD << endl;
    }
    return 0;
}

——written by Lyon