HDU 2199 (13.08.07)

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!
此题用二分法求解～
AC代码：
#include<stdio.h>double f(double x) {    return (8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6);}double myabs(double x) {    if(x < 0)        return x * (-1);    else        return x;}int main() {    int T;    scanf("%d", &T);    while(T--) {        double begin = 0.0;        double end = 100.0;        double mid;        double Y;        scanf("%lf", &Y);        if(f(0.0) > Y || f(100.0) < Y) {            printf("No solution!\n");            continue;        }        mid = (begin + end) / 2;        while(myabs(f(mid) - Y) > 1e-5) {            if(f(mid) > Y) {                end = mid;                mid = (begin + end) / 2;            }            else if(f(mid) < Y){                begin = mid;                mid = (begin + end) / 2;            }        }        printf("%.4lf\n", mid);    }    return 0;}