Can you solve this equation?（二分查找的简单应用）

Can you solve this equation?

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

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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

2100-4

 

Sample Output

1.6152No solution!

 思路：
记f(x)为等式左边的值，如果f(100.0)小于Y或f(0.0)大于Y，无解，否则必有解，且解的范围为[0.0, 100.0],不断二分区间，直至找到f(x) = Y

#include <iostream>#include <string>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <map>#include <vector>#include <queue>#include <stack>#define LL long long#define MAXI 2147483647#define MAXL 9223372036854775807#define eps (1e-8)#define dg(i) cout << "*" << i << endl;using namespace std;double Solve(double x){    return (8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6);}int main(){    int t;    double y, low, high, mid;    scanf("%d", &t);    while(t--)    {        scanf("%lf",&y);        low = 0.0;        high = 100.0;        if(y > Solve(100.0) || y < Solve(0.0)) puts("No solution!");        else        {            while(low + eps < high)            {                mid = (low + high) * 0.5;                if(y > Solve(mid) + eps) low = mid;                else if(y < Solve(mid) - eps) high = mid;                else break;            }            printf("%.4lf\n", mid);        }    }    return 0;}