hdu 2199

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Problem Description
Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can you find its solution between 0 and 100;
Now please try your lucky.

Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has a real number Y (fabs(Y) <= 1e10);

Output
For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100.

Sample Input
2
100
-4

Sample Output
1.6152

No solution!

迭代法代码：

#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <iomanip>#include <cmath>#include <algorithm>using namespace std;double cmp(double low,double high,double y);double mcp(double x);int main(){int T;double n,m;while(cin>>T){while(T--){cin>>n;if(n<mcp(0)||n>mcp(100))  cout<<"No solution!"<<endl;else {      m=cmp(0,100,n);      cout<<setiosflags(ios::fixed)<<setprecision(4)<<m<<endl;}          }}return 0;}double cmp(double low,double high,double y){double mid;while(high-low>1e-10){mid=(low+high)/2.0;if(mcp(mid)>y) high=mid;else low=mid;}return mid;}double mcp(double x){return 8.0*x*x*x*x+7.0*x*x*x+2.0*x*x+3.0*x+6.0;}

递归法代码：

如果递归的判断终止条件为low<high，时间超限；因为这个类型是double型，有些数会递归很多次以至n次才会的到结果，以至超时；由于提上要求是4位小数，所以只需要让low<high-0.0000001即可；

#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <iomanip>#include <cmath>#include <algorithm>using namespace std;double cmp(double low,double high,double y);double mcp(double x);int main(){int T;double n,m;while(cin>>T){while(T--){cin>>n;if(n<mcp(0)||n>mcp(100))  cout<<"No solution!"<<endl;else {      m=cmp(0,100,n);      cout<<setiosflags(ios::fixed)<<setprecision(4)<<m<<endl;}          }}return 0;}double cmp(double low,double high,double y){double mid=(low+high)/2;if(high-low>1e-10){    if(mcp(mid)>y)        return cmp(low,mid,y);        else if(mcp(mid)<y)            return cmp(mid,high,y);        else return mid;}return mid;}double mcp(double x){return 8.0*x*x*x*x+7.0*x*x*x+2.0*x*x+3.0*x+6.0;}

思路：先要判断这个函数的单调性，之后在就容易多了，给你一个数y；求解出对应的x，如果x是在0-100之间输出x浮点数为4；否则输出“No solution!”；

说白了就是把y移到等式左边是f(x)= 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 -Y；求是零点的值；如果x是在0-100之间输出x，浮点数为4；否则输出“No solution!”；

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