UVA 10341 Solve It （解方程 二分查找+精度）

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

题目大意：给你p,q,r,s,t,u六个变量的数值，要求你求出在区间（1,0）内，求x的值。

解析：对函数求导后导函数>0，可得函数在（1,0）上是单调递增的，所以直接用二分查找法求解。

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