Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6176 Accepted Submission(s): 2889
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
Sample Output
1.6152
No solution!
思路：浮点型的二分法
代码：
#include<iostream>
#include<stdio.h>
using namespace std;
double y;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lf",&y);
if(6<=y&&y<=807020306)
{
double m,l,r;
l=0;
r=100;
while(r-l>1e-6)
{
m=(l+r)/2;
if(8*m*m*m*m+7*m*m*m+2*m*m+3*m+6<y)
l=m+1e-7;
else
r=m-1e-7;
}
printf("%.4lf ",(l+r)/2.0);
}
else
printf("No solution! ");
}
return 0;
}
#include<stdio.h>
using namespace std;
double y;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lf",&y);
if(6<=y&&y<=807020306)
{
double m,l,r;
l=0;
r=100;
while(r-l>1e-6)
{
m=(l+r)/2;
if(8*m*m*m*m+7*m*m*m+2*m*m+3*m+6<y)
l=m+1e-7;
else
r=m-1e-7;
}
printf("%.4lf ",(l+r)/2.0);
}
else
printf("No solution! ");
}
return 0;
}