Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27728    Accepted Submission(s): 11717

 

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


【题意】

在[0,100]内找一个方程的实数解。


【分析】

求导易得函数单增,只需二分即可。

注意精度!


【代码】

 

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
#define debug(x) cerr<<#x<<" "<<x<<'\n';
using namespace std;
inline int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
const int N=1e5+5;
int n;double Y;
inline double f(double x){
return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;
} int main(){
for(n=read();n--;){
scanf("%lf",&Y);
if(Y<f(0)||Y>f(100)){puts("No solution!");continue;}
double l=0,r=100,mid,ans;
while(r-l>1e-6){
mid=(l+r)/2.0;
if(Y-f(mid)<=1e-5){
ans=mid;
r=mid;
}
else{
l=mid;
}
}
printf("%.4lf\n",ans);
}
return 0;
}

 

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