uva 10341 - Solve It

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 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

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

#include<string.h>#include<stdio.h>#include<math.h>double p,q,r,s,t,u;double ans(double x){return p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u;};double answer,f;void find(double left,double right){ double mid=(left+right)/2,k1,k2,k3; if ((f>0)||(left-right>=1e-6)) return; k1=ans(left); k2=ans(mid); k3=ans(right); if (k1*k3>0) return ; if (fabs(k1)<1e-6) {answer=left;f=1;return;} if (fabs(k3)<1e-6) {answer=right;f=1;return;} if (fabs(k2)<1e-6) {answer=mid;f=1;return;} if (k1*k2<0) find(left,mid); if (k2*k3<0) find(mid,right);}int main(){ while (scanf("%lf%lf%lf%lf%lf%lf",&p,&q,&r,&s,&t,&u)!=EOF) {  f=-1;  find(0,1);  if (f>0) printf("%.4lf\n",answer);    else printf("No solution\n"); } return 0;}