UVA 10341 (二分查找+精度)

题意：

给你一个关于x的方程，给出变量的值，求出x；

Problem F

Solve It

Input:standard input

Output:standard output

Time Limit: 1 second

Memory Limit: 32 MB

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,t and u (where0 <= p,r <= 20 and-20 <=q,s,t <= 0). There will be maximum 2100 lines in the input file.

Output

For each set of input, there should be a line containing the value ofx, correct upto 4 decimal places, or the string "No solution", whichever is applicable.

Sample Input

0 0 0 0 -2 1

1 0 0 0 -1 2

1 -1 1 -1 -1 1

Sample Output

0.7071

No solution

思路：
二分法零点定理找根，必须在[0,1]上单调，求导f'(x)=-p*e(-x)+q*cos(x)-r*sin(x)+s/(cos(x)*cos(x))+2t*x
x      , [0,1]
e（-x） ，[1/e，1]
cos（x），[0,pi]
sin(x) , [0,pi]
p,r>0 q,s,t<0, f'(x)的每一项都是负的，
f（x）单调递减,故二分枚举，当f(left) *f(right)>0,无解

代码：

#include <iostream>// 二分查找，#include <algorithm>#include <string>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <stack>#include <map>#include <set>using namespace std;typedef long long ll;typedef pair<int,int> P;const int maxn=105;const int base=1000;const int inf=0x3f3f3f3f;const double eps=1e-8;const double pi=acos(-1.0);double p,q,r,s,t,u;double fun(double x){    return p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*pow(x,2)+u;}int main(){    int n,m,i,j;    while(~scanf("%lf%lf%lf%lf%lf%lf",&p,&q,&r,&s,&t,&u))    {        int sum=0;        double left=0, right=1, mid;        bool flag=false;        if(fun(left)*fun(right)>0)  //在同一侧，无解        {            printf("No solution\n");            continue;        }         while(right-left>eps)      //二分逼近        {            mid=(left+right)/2;            if(fun(mid)*fun(left)>0) left=mid;            else right=mid;        }        printf("%.4f\n",mid);    }    return 0;}