UVA 10341 (13.11.07)

Problem F

Solve It

Input: standard input

Output: standardoutput

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 testcase consists of 6 integers in a single line:p,q, r, s, tand 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

题目给出的公式是一个单调递减函数...

然后找值用二分...你们懂的不多说

做简单题也有收获, 例如提醒了自己fabs()的使用~

AC代码:

#include<stdio.h>  
#include<string.h>  
#include<math.h>  
  
double p, q, r, s, t, u;  
  
double getAns(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) {  
        int mark = 1;  
        double l = 0;  
        double r = 1;  
        while(fabs(l-r) > 1e-10) {  
            if((getAns(l) > 0 && getAns(r) > 0) || (getAns(l) < 0 && getAns(r) < 0)) {  
                printf("No solution\n");  
                mark = 0;  
                break;  
            }  
            double t = (l + r) / 2;  
            if(getAns(t) > 0)  
                l = t;  
            if(getAns(t) < 0)  
                r = t;  
        }  
        if(mark)  
            printf("%.4lf\n", r);  
    }  
    return 0;  
}

补充：软件开发 , C++ ,