UVa 10341 Solve It (牛顿法解超越方程)

10341 - Solve It

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_problem&problem=1282

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

参考：维基百科——牛顿法

注意：

1. 根据系数的范围知道方程左端是递减的。（可以使用零点定理）

2. 如果f'(x)不为0, 那么牛顿法将具有平方收敛的性能.。粗略的说，这意味着每迭代一次，牛顿法结果的有效数字将增加一倍。

PS：你也可以用二分法，只不过慢一点而已。

完整代码：

/*0.019s*/#include <cstdio>#include <cmath>int p, q, r, s, t, u;double x;inline double fun(double x){return p * exp(-x) + q * sin(x) + r * cos(x) + s * tan(x) + t * x * x + u;}///导数inline double fun2(double x){return -p * exp(-x) + q * cos(x) - r * sin(x) + s / (cos(x) * cos(x)) + 2 * t * x;}int main(void){while (~scanf("%d%d%d%d%d%d", &p, &q, &r, &s, &t, &u)){double x1 = fun(0.0), x2 = fun(1.0);if (x1 == 0){puts("0.0000");continue;}if (x2 == 0){puts("1.0000");continue;}if (x1 * x2 > 0){puts("No solution");continue;}x = 0.5;///从0.5开始double tmp = 0.0;while (fabs(x - tmp) > 1e-3)///想保险点的话可以写1e-6(保两位){tmp = x;x -= fun(x) / fun2(x);}printf("%.4f\n", x);}return 0;}