(HDU

(HDU - 2199)Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 20985 Accepted Submission(s): 9208

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!

题目大意：已知方程8∗x4+7∗x3+2∗x2+3∗x+6=y，给定一个y让你求在区间[0,100]的解。

思路：显然这个方程是单调递增的，所以可以二分来做。显然当y<6||y>807020306没有解；上式等价为求解8∗x4+7∗x3+2∗x2+3∗x+6−y=0，已知在区间[0,100]二分答案即可。

#include<cstdio>using namespace std;const double eps=1e-8;double y;double f(double x){    return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6-y;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%lf",&y);        if(y<6||y>807020306) printf("No solutions!\n");        else        {            double lo=0,hi=100,mid;            while(hi-lo>eps)            {                mid=(lo+hi)/2.0;                if(f(mid)>0) hi=mid;                else lo=mid;             }            printf("%.4f\n",mid);        }    }    return 0;}