二分法、三分法

//http://acm.hdu.edu.cn/showproblem.php?pid=2675//Equation Again/*题目等价于 x^y = y^xln(x)/x = ln(y)/y所以 x和y是f(z) = ln(z)/z函数取某个值时的不同的z的两个自变量再打一下表 汇出f(z)的图像显然x=y(y>=e)是它的一个解，由图像还有一个是在1和e之间所以用二分法查找 得解*/#include<stdio.h>#include<math.h>double f(double x){return log(x)/x;}const double e=2.718281828459045;int main(){double y;while(scanf("%lf",&y)!=EOF){double w=f(e*y);double head=1,rear=e;double x=(head+rear)/2;while(fabs(rear-head)>=1e-7){if(f(x)<w)head=x;if(f(x)>w)rear=x;x=(head+rear)/2;//printf("%lf %lf/t",head,rear);}if(y==1)printf("%.5lf/n",e*y);elseprintf("%.5lf %.5lf/n",x,y*e);//printf("%lf/n",w);}return 0;}