Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12794 Accepted Submission(s): 5715
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!
本题的大致意思是：8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y，给出一个Y，算出的X如果在0到100之间，就输出，否则就输出No solution!
题解：题目很简单，但是一开始根本没有思路。此时博主在学二分查找专题，就往二分查找上面想，自己也想不出来。看了大神题解才知道，可以二分x，下限为0，上限为100.用软件画了这个函数，它在0到100之间是单调的，当x等于0最小，Y等于6，当x等于100最大，Y等于807020306.所以等Y<6或者Y>807020306时，输出No solution!如果Y值在这之间，就二分x，在0到100之间，如果x带入公式比Y小，那就放大x，反之就放小x。结束的条件为上限减去下限不大于1e-10（最小数）.
误解：将x带入公式算出来的值，很少有情况会等于Y，或者说在double精确范围下，根本不会使得带入x算出的值等于Y，只能让算出的值无限接近Y，那这个x就是答案（x精确度为1e-10）。
具体代码如下：
#include<iostream>
#include<cmath>
#include<cstdio>
int t=0;
using namespace std;
double Search(double y,double low,double high)//在【0,100】内二分x
{
double mid=(high+low)/2.0;
if(high-low>1e-8)
{
if(8.0*pow(mid,4)+7.0*pow(mid,3)+2.0*pow(mid,2)+3.0*mid+6.0>y)
return Search(y,low,mid);
return Search(y,mid,high);
}
return mid;
}
int main()
{
int n;
cin>>n;
while(n--)
{
double y;
cin>>y;
t=0;
if(y<6||y>807020306)
printf("No solution!\n");
else
printf("%.4f\n",Search(y,0,100));
}
}