UVA Solve It（二分查找）

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：找到一个数X满足 p*e-x+ q*sin(x) + r*cos(x) + s*tan(x) + t*x2 + u = 0

代码：

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>using namespace std;double p,q,r,s,t,u;double findx(double x){    return p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u;}int main(){    while(scanf("%lf%lf%lf%lf%lf%lf",&p,&q,&r,&s,&t,&u)!=EOF)    {        if(findx(0)<0 || findx(1)>0)        {            printf("No solution\n");        }        else        {            double x1 = 0;            double x2 = 1;            while(fabs(x1-x2)>=1e-10)            {                double x = (x1+x2)/2;                if(findx(x)>0)                {                    x1 = x;                }                else                {                    x2 = x;                }            }            printf("%.4lf\n",x1);        }    }    return 0;}

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Mustaq Ahmed