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;
}