UVA 10341- Solve It(二分+精度求解方程值)

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 (where 0 <= 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 of x, 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

0.7554

题意：给出p,q,r,s,t,u的值，根据方程式求出方程的近似解。

思路：因为方程是递减的，对于两个端点，如果f(0)<0或者是f(1)>0 ,则说明方程无解。取两个端点的中点mid 如果f(0)*f(mid)<0  ,说明方程的解在[0,mid]中，否则解在[mid,1]中，这样不断进行下去，区间会收敛于一点，即为最后的近似解

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
#define eps 1e-8
double p,q,r,s,t,u;
double Solve (double x)
{
    return p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u;//exp(x) 指的是e的x次方。
}
int main()
{
    double left,right,mid;
    while(~scanf("%lf %lf %lf %lf %lf %lf",&p,&q,&r,&s,&t,&u)) {
        if(Solve(0)<0||Solve(1)>0)
            printf("No solution\n");
        else {
            left=0;
            right=1;
            while((right-left)>=eps) {
                mid=(left+right)/2;
                if(Solve(mid)<0)
                    right=mid;
                else
                    left=mid;
            }
            printf("%.4lf\n",mid);
        }
    }
    return 0;
}