UVA 10341 solve it

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 (where 0 ≤ 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 of x, correct up to 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

观察题目中的条件，方程必定在定义域内单调递减，逻辑方面就很容易了。

在实现上，使用二分找到一个满足条件的解，不过我一开始使用的whlie循环居然会超时，后来换成for循环就过了，后来才发现我的循环有问题

while(abs(f(mid))>=MIN)
            {
                if(f(mid)>=MIN)
                fir=mid,mid=(fir+las)/2;
                else
                las=mid,mid=(fir+las)/2;
        }

判断结束的精确度不应该是用函数值，而应该是用mid，下面for循环通过代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#include<map>
using namespace std;
const int MAX=1e6;
const double MIN=1e-6;
#define INF 0x7fffffff
#define ll long long
#define FOR(i,n) for(i=1;i<=n;i++)
#define mst(a) memset(a,0,sizeof(a))
#define mstn(a,n) memset(a,n,sizeof(a))
//struct num{int a,b,i;}a[1005];
//bool cmp(const num &x, const num &y){return x.a>y.a;}
int  p,q,r,s,t,u;
float f(float x)
{
    return p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u;
}

int main()
{
    while(scanf("%d%d%d%d%d%d",&p,&q,&r,&s,&t,&u)!=EOF)
    {
        double fir=0,las=1,mid=(fir+las)/2;
        if(f(fir)<0||f(las)>0) 
        cout<<"No solution"<<endl;
        else
        {
            for(int i=0;i<=100;i++)
            {
                mid=(las+fir)/2;
                if(f(mid)>MIN)
                fir=mid;
                else
                las=mid;
            }
            printf("%.4lf
",mid);
     　}
　　}
    return 0;
}