Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23060 Accepted Submission(s):
10003
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!
Author
Redow
Recommend
这是一道比较基础的纯粹二分算法的的题目。 这一题要求的就是8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y，给定Y在0-100内找出一个数X是这个式子成立，当然也就是找出一个double数字使左边和右边最相近。我是按照二分，然后判断mid带入上面式子之后比10^-8小，但是经过实践是不行的，可以过案例但是会ＷＡ，我不是很明白可能是精度的问题，所以不要这样做。正确的做法是：从 low＝0和high=100往中间挤，也就是当low和high的差很小时，mid的差别也很小，那么mid代入上式和Y的差距也就小了，实践证明开到e-8，e-7都是可以的。
1 #include <iostream> 2 #include <cstring> 3 #include <string> 4 #include <algorithm> 5 #include <cmath> 6 #define eps 1e-6 7 using namespace std; 8 double A(double x) 9 { 10 return (8 * pow(x, 4) + 7 * pow(x, 3) + 2 * pow(x, 2) + 3 * pow(x, 1) + 6); 11 } 12 int main() 13 { 14 int n; 15 long long y, t; 16 cin >> n; 17 while (n--) 18 { 19 cin >> y; 20 if (A(0)>y || A(100)<y) 21 { 22 cout << "No solution!" << endl; 23 continue; 24 } 25 if (y == 0) cout << 0 << endl; 26 else 27 { 28 double l = 0, r = 100; 29 double mid; 30 while (r - l >=eps) 31 { 32 mid = (l + r) / 2; 33 if (A(mid) > y) 34 { 35 r = mid; 36 } 37 else l = mid; 38 } 39 printf("%.4lf ", mid); 40 } 41 } 42 return 0; 43 }