1.题目描述：点击打开链接
2.解题思路：本题要求解一个超越方程，通过观察发现，方程的前五项在区间[0,1]上单调递减，因此如果端点F(0)≤0或F(1)≥0，则无解。否则，一定有解。此时可以利用二分搜素来求出解。由于实数区间可以无限二分，因此通常是当区间长度小于特定值eps时终止，例如eps=1e-5；也可以当迭代次数超过一定值时终止，例如100次。
3.代码：
#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;int p, q, r, s, t, u;double F(double x){return p*exp(-x) + q*sin(x) + r*cos(x) + s*tan(x) + t*x*x + u;}const double eps = 1e-14;int main(){//freopen("t.txt", "r", stdin);while (~scanf("%d%d%d%d%d%d", &p, &q, &r, &s, &t, &u)){double f0 = F(0), f1 = F(1);if (f1 > eps || f0 < -eps)printf("No solution\n");else{double x = 0, y = 1, m;for (int i = 0; i < 100; i++)//限制迭代次数{m = x + (y – x) / 2;if (F(m) < 0)y = m;else x = m;}printf("%.4lf\n", m);}}return 0;}
，感受最美的风景。你曾经说，
