hdu 2199 Can you solve this equation? にぶん
7541 ワード
Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6763 Accepted Submission(s): 3154
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!
Author
Redow
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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6763 Accepted Submission(s): 3154
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!
Author
Redow
Recommend
lcy | We have carefully selected several similar problems for you:
2899
2289
2298
2141
3400
1 /*
2 , 。
3 , 4 , 1e-5 。
4
5 */
6
7 #include<iostream>
8 #include<stdio.h>
9 #include<cstring>
10 #include<cstdlib>
11 #include<math.h>
12 using namespace std;
13
14 double fun(double x)
15 {
16 return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;
17 }
18 void EF(double l,double r,double Y)
19 {
20 double mid;
21 while(r-l>1e-7)
22 {
23 mid=(l+r)/2;
24 double ans=fun(mid);
25 if( ans >Y )
26 r=mid-1e-8;
27 else l=mid+1e-8;
28 }
29 printf("%.4lf
",(l+r)/2);
30 }
31 int main()
32 {
33 int T;
34 double Y;
35 scanf("%d",&T);
36 {
37 while(T--)
38 {
39 scanf("%lf",&Y);
40 if( fun(0.0)>Y || fun(100.0)<Y)
41 printf("No solution!
");
42 else
43 EF(0.0,100.0,Y);
44 }
45 }
46 return 0;
47 }