Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7689    Accepted Submission(s): 3574

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!

 
方程单调，二分

#include<stdio.h>
#include<math.h>
double cal(double x){
     return 8*pow(x,4) + 7*pow(x,3) + 2*pow(x,2) + 3*x + 6;
}
int main(){
    //freopen("in.txt","r",stdin);
    int zu;
    while(scanf("%d",&zu)!=EOF){
        while(zu--){
            double y;
            scanf("%lf",&y);
            if(y>cal(100)||y<cal(0)){
                printf("No solution!\n");
                //不是break;，因为后面还有数字要处理
                continue;
            }
            double start=0,end=100,mid;
            //1e-6代表10的负6次方
            //判断浮点数的大小，是用2数相减，得到的差和一个非常小的数字比较
             while( end -start     > 1e-6 ) {
                mid = (start + end) / 2;
                double ans = cal(mid);
                if( ans > y ) {
                    end=mid;
                }else{
                    start = mid;
                }
                    
            }
             printf("%.4lf\n", (start + end) / 2 );

            
        }
    
    }
    return 0;
}