hdu 2199 Can you solve this equation?

最新推荐文章于 2019-08-13 19:42:15 发布

最新推荐文章于 2019-08-13 19:42:15 发布 · 143 阅读

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#java

本文探讨了如何求解在特定区间内（0到100）的复杂四次方程，并通过实例展示了求解过程及注意事项。

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Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2458Accepted Submission(s): 1234

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!

#include<stdio.h>
#include<math.h>
double y;
int flag;
double find(double mid)
{
double ans;
ans=8*mid*mid*mid*mid+7*mid*mid*mid+2*mid*mid+3*mid+6;
return ans;
}
int main()
{
 int cas;
double mid,start,end; 
scanf("%d",&cas);
while(cas--)
{
scanf("%lf",&y);
flag=0;
if(find(0)<=y&&find(100)>=y)//这个才是判断是否存在的关键条件
{
start=0;end=100;
while(end-start>0.0000001)//这个地方要尽量小 当为10的负6次方 仍旧不能达到样例数据中的精度 但是能AC
{ //所以以后做关于精度的题目 尽量小些 反正也不会对题目产生什么大影响
mid=(start+end)/2;
if(find(mid)>y) end=mid;
else //if(find(mid)<y)
start=mid;
}
printf("%.4lf\n",(start+end)/2);
}
else printf("No solution!\n");
}
return 0;
}