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  • UVA 10341

                                                                                        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 }
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  • 原文地址:https://www.cnblogs.com/Aa1039510121/p/5689669.html
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