hdu - 2199 - Can you solve this equation?（二分）

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题意：求8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y在[0, 100]上的解。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2199

——>>简单二分。

发现，在hust上用函数写法0ms过，用define写法15ms过，而这两种写法在hdu上都是0ms过。奇葩～

函数写法：

#include <cstdio>#include <cmath>using namespace std;const double eps = 1e-14;double Y;double F(double x){    return 8*pow(x, 4) + 7*pow(x, 3) + 2*pow(x, 2) + 3*x + 6 - Y;}int main(){    int T, i;    scanf("%d", &T);    while(T--)    {        scanf("%lf", &Y);        double f0 = F(0), f100 = F(100);        if(f0 > eps || f100 < -eps) printf("No solution!\n");        else        {            double x = 0, y = 100, m;            for(i = 0; i < 100; i++)            {                m = x + (y - x) / 2;                if(F(m) < 0) x = m;                else y = m;            }            printf("%.4lf\n", m);        }    }    return 0;}

define写法：

#include <cstdio>#include <cmath>#define F(x) (8*pow(x, 4) + 7*pow(x, 3) + 2*pow(x, 2) + 3*x + 6 - Y)using namespace std;const double eps = 1e-14;double Y;int main(){    int T, i;    scanf("%d", &T);    while(T--)    {        scanf("%lf", &Y);        double f0 = F(0), f100 = F(100);        if(f0 > eps || f100 < -eps) printf("No solution!\n");        else        {            double x = 0, y = 100, m;            for(i = 0; i < 100; i++)            {                m = x + (y - x) / 2;                if(F(m) < 0) x = m;                else y = m;            }            printf("%.4lf\n", m);        }    }    return 0;}