HDU 2899(模拟退火做法)

地址：​​http://acm.hdu.edu.cn/showproblem.php?pid=2899​​

Now, here is a fuction:
F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.

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 only one real numbers Y.(0 < Y <1e10)

Output

Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.

#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<stack>
#include<queue>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const double eps=1e-8;　　//终止温度
double y;　　　　
double func(double x)
{
    return 6*pow(x,7.0)+8*pow(x,6.0)+7*pow(x,3.0)+5*pow(x,2.0)-x*y;
}
double solve()
{
    double T=100;　　//初始温度
    double delta=0.98;//降温系数
    double x=50.0;　　//x的初始值
    double now=func(x);　　　　//初始函数值
    double ans=now;　　　　//记录结果
    while(T>eps)
    {
        int f[2]={1,-1};
        double newx=x+f[rand()%2]*T;　　//按概率改变x
        if(newx>=0&&newx<=100)
        {
            double next=func(newx);
            ans=min(ans,next);
            if(now-next>eps)
            {
                x=newx;
                now=next;
            }
        //    cout<<x<<endl;
        }
        T*=delta;
    }
    return ans;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf",&y);
        printf("%.4lf\n",solve());
    }
}