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…

  Three Families  Three families share a garden. They usually clean the garden together at the end of each week, but last week, family C was on holiday, so family A spent 5 hours, family B spent 4 hours and had everything done. After coming back, fam…

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 <…

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…

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.    …

There is a war and it doesn't look very promising for your country. Now it's time to act. You have a commando squad at your disposal and planning an ambush on an important enemy camp located nearby. You have N soldiers in your squad. In your master-p…

Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predators, the geese population was out o…

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…

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…

UVa 489 题目大意:计算机给定一个单词让你猜,你猜一个字母,若单词中存在你猜测的字母,则会显示出来,否则算出错, 你最多只能出错7次(第6次错还能继续猜,第7次错就算你失败),另注意猜一个已经猜过的单词也算出错, 给定计算机的单词以及猜测序列,判断玩家赢了(You win).输了(You lose).放弃了(You chickened out) /* UVa 489 HangmanJudge --- 水题 */ #include <cstdio> #include <cstring…