题意:给出一个式子以及里面的常量,求出范围为[0,1]的解,精度要求为小数点后4为. 二分暴力查找即可. e^(-n)可以用math.h里面的exp(-n)表示. 代码:(uva该题我老是出现Submission Error,过几天再试看看) /* * Author: illuz <iilluzen@gmail.com> * Blog: http://blog.csdn.net/hcbbt * File: uva10241.cpp * Lauguage: C/C++ * Create Date…

很水的一道题,因为你发现这个函数是单调递减的,所以二分法求出函数的根即可. #include <cstdio> #include <cmath> //using namespace std; ; double p, q, r, s, t, u; inline double f(double x) { return p*exp(-x) + q*sin(x) + r*cos(x) + s*tan(x) + t*x*x + u; } int main() { //freopen(&quo…

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu   Description   Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber…

题目:给一个方程,求解方程的解.已给出解的范围,并且可知方程等号左侧的函数是递减的,可用二分法进行试探,直到得出给定误差范围内的解. #include <cstdio> #include <cmath> #define EPSILON 1e-9 int p, q, r, s, t, u; double f(double x) { return p*exp(-1.0*x) + q*sin(x) + r*cos(x) + s*tan(x) + t*x*x + u; } int main…

原题: Solve the equation:         p*e-x + q*sin(x) + r*cos(x) + s*tan(x) + t*x2 + u = 0         where 0 <= x <= 1. Input Input consists of multiple test cases and terminated by an EOF. Each test case consists of 6 integers in a single line: p, q, r, s…

题目大意:给6个系数,问是否存在X使得等式成立 思路:二分.... #include <stdio.h> #include <math.h> #define EEE 2.71828182845953581496 int p, q, r, s, t, u; double v(double x) { return (p*pow(EEE,-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x); } double f(double x,double z, double…

Solve the equation:p ∗ e−x + q ∗ sin(x) + r ∗ cos(x) + s ∗ tan(x) + t ∗ x2 + u = 0where 0 ≤ x ≤ 1.InputInput consists of multiple test cases and terminated by an EOF. Each test case consists of 6 integers ina single line: p, q, r, s, t and u (where 0…

题目链接 题意: 解方程:p ∗ e^(−x) + q ∗ sin(x) + r ∗ cos(x) + s ∗ tan(x) + t ∗ x^2 + u = 0 (0 <= x <= 1); 其中0 ≤ p, r ≤ 20 , −20 ≤ q, s, t ≤ 0.(一开始没看见q,s,t<=0, 卡了半天...) 根据上面的条件,设F(x) = p ∗ e^(−x) + q ∗ sin(x) + r ∗ cos(x) + s ∗ tan(x) + t ∗ x^2 + u ;即求0 <…

Solve the equation:p ∗ e ^−x + q ∗ sin(x) + r ∗ cos(x) + s ∗ tan(x) + t ∗ x ^2 + u = 02 + u = 0where 0 ≤ x ≤ 1.InputInput consists of multiple test cases and terminated by an EOF. Each test case consists of 6 integers ina single line: p, q, r, s, t a…

二分查找 二分查找又称折半查找,优点是比较次数少,查找速度快,平均性能好:其缺点是要求待查表为有序表,且插入删除困难.因此,折半查找方法适用于不经常变动而查找频繁的有序列表.首先,假设表中元素是按升序排列,将表中间位置记录的关键字与查找关键字比较,如果两者相等,则查找成功:否则利用中间位置记录将表分成前.后两个子表,如果中间位置记录的关键字大于查找关键字,则进一步查找前一子表,否则进一步查找后一子表.重复以上过程,直到找到满足条件的记录,使查找成功,或直到子表不存在为止,此时查找不成功.    …