POJ 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

0.7554

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Mustaq Ahmed

    题目中的x是代表的弧度，而不是度数。 可以研究一下他的单调性，很容易就得出他是一个单调递减的函数。(提示：分别研究每一个小函数，就会得出，每一个小函数都是递减的，那么整个函数就是递减的) 然后剩下的就是用二分就可以了。要注意精度啊。 只是要到1e-9

#include <stdio.h>#include <string.h>#include <math.h>#define EQS 1e-9double p,q,r,s,t,u;int k;int main(){    double binary_search(double l,double r);    int i,j,n,m;    double x,res;    while(scanf("%lf %lf %lf %lf %lf %lf",&p,&q,&r,&s,&t,&u)!=EOF)    {        k=0;        u=u*-1;        res=binary_search(0.0,1.0);        if(k==1)        {            printf("%.4lf\n",res);        }else        {            printf("No solution\n");        }    }    return 0;}int check(double x){    double check_s;    check_s= p*exp(-1*x) + q*sin(x) + r*cos(x) + s*tan(x) + t*x*x;    if(fabs(check_s-u)<=1e-7)    {        return 0;    }else if(check_s>u)    {        return 1;    }else    {        return -1;    }}double binary_search(double l,double r){    double mid;    int t;    while(!(r<l||fabs(l-r)<=EQS))    {        mid= (l+r)/2.0;        t=check(mid);        if(t==0)        {            k=1;            return mid;        }else if(t==-1)        {            r=mid;        }else        {            l=mid;        }    }    return -1;}