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 ofx, 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
函数单调递减 二分查找，高中知识，区间头尾对应的值相乘为负数，开始我分左右两边，用二分查找找他们相等，虽然给的样例过了，
但一直wa，也许是精度缺失了把，最后选择这样的
1 #include<iostream> 2 #include<cmath> 3 #include<cstdio> 4 using namespace std; 5 int p,q,r,s,t,u; 6 double fun(double x) 7 { 8 double y; 9 y = p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u; 10 return y; 11 } 12 int main() 13 { 14 double m,n; 15 while(cin>>p>>q>>r>>s>>t>>u) 16 { 17 double max=1.0; 18 double min=0.0; 19 if(fun(max)*fun(min)>0) 20 cout<<"No solution"<<endl; 21 else 22 { 23 double mid; 24 while(min+0.00000001<=max) 25 { 26 mid=(max+min)/2.0; 27 if(fun(mid)<=0) 28 max=mid; 29 else 30 min=mid; 31 } 32 printf("%.4lf ",mid); 33 } 34 } 35 return 0; 36 }