Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25633 Accepted Submission(s): 11018
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!
题意：知道y值，要求算出x。并且x只能在0~100之间，不然就输出No solution!
题解：因为x只能在0~100之间。所以在知道y值得情况下判断有没有解的方法是：如果给出的Y值比f(0)还小，那他肯定没有0~100之间的解。因为解在0~100之间都是正数。同理也不能大于f(100);
其实上面的可以用这个函数在0~100之间单调递增来解释，比较清楚
剩下的就是二分来找解，看一下代码还是挺容易理解的
1 #include<bits/stdc++.h> 2 using namespace std; 3 double f(double x) 4 { 5 return (8*x*x*x*x+7*x*x*x+2*x*x+3*x+6); 6 } 7 int main() 8 { 9 10 int t; 11 while(~scanf("%d",&t)) 12 { 13 while(t--) 14 { 15 double y; 16 scanf("%lf",&y); 17 if(f(0)>y||f(100)<y) 18 { 19 printf("No solution! "); 20 continue; 21 } 22 double l,r; 23 l=0.0;r=100.0; 24 double mid=50.0; 25 while(fabs(f(mid)-y)>1e-5) 26 { 27 if(f(mid)>y) 28 { 29 r=mid; 30 mid=(l+r)/2.0; 31 32 } 33 else 34 { 35 l=mid; 36 mid=(l+r)/2.0; 37 } 38 39 } 40 printf("%.4lf ",mid); 41 } 42 } 43 return 0; 44 }