UVa10341

10341 Solve It
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 up to 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：

       解方程p*e^(-x)+q*sin(x)+r*cos(x)+s*tan(x)+tx^2+u=0，其中0<=x<=1。0<=p,r<=20

-20<=q,s,t<=0。

分析：

       设为F(x)。注意到F(x)在[0,1]内单调递减，用二分即可解。

#include <stdio.h>
#include <math.h>
#define F(x) (p * exp(-x) + q * sin(x) + r * cos(x) + s * tan(x) + t * (x) * (x) + u)
const double eps = 1e-14;
int p,q,r,s,t,u;
int main(){
    while(scanf("%d%d%d%d%d%d",&p,&q,&r,&s,&t,&u) == 6){
        double f0 = F(0),f1 = F(1);
        if(f1 > eps || f0 < -eps) printf("No solution\n");
        else{
            double x = 0,y = 1,m;
            for(int i = 0 ; i < 100 ; i++){
                m = x + (y - x) / 2;
                if(F(m) < 0) y = m;
                else x = m;
            }
            printf("%.4lf\n",m);
        }
    }
    return 0;
}