Can you solve this equation?

二分，函数递增，所以可以先确定y的区间，不在区间则为No solution！

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!

#include<stdio.h>#include<math.h>bool judge(double n);double y,x;int main(){double le,ri,mid,max;int t,i;scanf("%d",&t);max=8*pow(100,4)+7*pow(100,3)+2*pow(100,2)+3*100;while(t--){scanf("%lf",&y);y-=6;if(y<0||y>max)printf("No solution!\n");else{le=0;ri=y;mid=(le+ri)/2;while(ri-le>=0.0000000001){if(judge(mid))le=mid;elseri=mid;mid=(le+ri)/2*1.0;}if(mid==0)printf("0\n");elseprintf("%.4lf\n",mid); }}}bool judge(double n){double sum;sum=8*pow(n,4)+7*pow(n,3)+2*pow(n,2)+3*n;if(sum>=y)return 0;elsereturn 1;}