Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16473 Accepted Submission(s): 7301
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!
这个题运用二分很好算，当然要注意函数是否递增，求一阶、二阶导数即可，注意判断条件，因为是double型的，所以要用1e-12.
代码如下：
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <string.h>
#define eps 1e-12
using namespace std;
double cmp(double x){
return 8*pow(x,4.0)+7*pow(x,3.0)+2*pow(x,2.0)+3*x+6;
}
int main(){
int n;
double y,z=8*pow(100.0,4.0)+7*pow(100.0,3.0)+2*pow(100.0,2.0)+306.0;
scanf("%d",&n);
while(n--){
scanf("%lf",&y);
double l=0,r=100;
if(y-6<eps||y>z){
printf("No solution!\n");
continue;
}
else{
int size=100;
while(size--){
double mid=(l+r)/2.0;
if(y>cmp(mid))
l=mid;
else
r=mid;
}
}
printf("%.4lf\n",l);
}
return 0;
}