HDU - 2199 Can you solve this equation? 二分 简单题
2024-08-27 07:10:53
Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22742 Accepted Submission(s): 9865
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.
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
100
-4
Sample Output
1.6152
No solution!
No solution!
二分求解
#include<stdio.h>
const double eps = 1e-8;//精度大胆设高些 double cal(double x,double y)
{
double m;
m = 8 * x*x*x*x + 7 * x*x*x + 2 * x*x + 3 * x + 6 - y;
return m;
} double bsearch(double y)
{
double left = 0, right = 100;
double mid;
while (right - left > eps)
{
mid = (left + right) / 2;
if (cal(mid, y)>0)
{
right = mid;
}
else
{
left = mid;
}
}
return mid;
} int main()
{
int T;
scanf("%d", &T);
double M,n;
while (T--)
{
scanf("%lf", &M);
if (M < 6 || M>807020306)//注意上下界
{
printf("No solution!\n");
}
else
{
n=bsearch(M);
printf("%.4lf\n", n);
}
}
return 0;
}
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