HDU-2199-Can you solve this equation?【二分】

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 16553    Accepted Submission(s): 7343

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

2100-4

 

Sample Output

1.6152No solution!

代码一：

#include<cstdio>#include<algorithm>#include<cmath>#define eps 1e-12using namespace std;double f(double x){return 8*pow(x,4.0)+7*pow(x,3.0)+2*pow(x,2)+3*x+6;}int main(){int t;scanf("%d",&t);while(t--){double y,mid,maxn;double l=0.0,r=100.0;scanf("%lf",&y);maxn=8*pow(100.0,4.0)+7*pow(100.0,3.0)+2*pow(100.0,2)+3*100+6;if(y<6||y>maxn){printf("No solution!\n");continue;}int size=50;while(size--)  //循环次数一定比 50 少 {mid=(l+r)/2.0;if(y<f(mid))r=mid-eps;  //不加 eps 这个精度也能 AC elsel=mid+eps;}printf("%.4lf\n",l);}return 0;}

代码二：

#include<cstdio>#include<cmath>double f(double x){return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;}int main(){int t,a;scanf("%d",&t);while(t--){double y,l,r,mid;scanf("%lf",&y);if(y < 6 || y > 8*100*100*100*100+7*100*100*100+2*100*100+3*100+6)printf("No solution!\n");else{l=0;r=100;a=50;mid=(l+r)/2;while(fabs(f(mid)-y)>1e-5){if(f(mid) < y){l=mid+1;}else{r=mid-1;}mid=(l+r)/2;}printf("%.4f\n",mid);}}}