简单二分 UVA 10341

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


这种解方程求解题，先考虑下二分，通过系数判定函数形势，是个递减的非线性方程，给出具体方程，要在[0,1]里找方程对应的x。
因为是取后四位，精度开到10e-5以上我觉得都是可以的，当然也要满足运行时间的要求。
因为是递减的，靠在0那边就大于0，靠1那边就小于0，用f*f的符号判断解的位置。

代码：

#include<bits/stdc++.h>
using namespace std;

double p,q,r,s,t,u;
double f(double x)
{
return p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u;
}
int main()
{
double l,r,m;

while(~scanf("%lf%lf%lf%lf%lf%lf",&p,&q,&r,&s,&t,&u))
{
l=0;r=1;
if(f(l)*f(r)>0) {cout<<"No solution"<<endl; continue;}
while(r-l>10e-6)
{
m=(r+l)/2;
if(f(l)*f(m)>0)
l=m;
else r=m;
}
printf("%.4lf\n", m);
}
return 0;
}