Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 25430    Accepted Submission(s): 10944  

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!

題意是：

輸入一個Y，你找出一個x,滿足8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,的區間為【0，100】

思路：二分搜尋，通過求導，可知該函式在【0，100】間是遞增的，首先先判斷Y是否能求出x

#include<stdio.h>
#include<math.h>
const double eps = 1e-6;

double f(double x)
{
   return 8*pow(x,4) + 7*pow(x,3) + 2*pow(x,2) + 3*x + 6 ;	
}

int main()
{
	int t;
	double l,r,mid,n;
	double ans=0.0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lf",&n);
		if(f(0)<=n&&n<=f(100))  //判斷n是否在此區間中
		{
			l=0,r=100;
			while(r-l>1e-6)
			{
				mid=(l+r)/2;
				ans = f(mid);
			    if(fabs(ans-n)<=eps) break;
				else if(ans > n) r=mid-1e-7;
				else l=mid+1e-7;
			}
			printf("%.4lf\n",mid);
		}
		else
		   printf("No solution!\n");
	}
	return 0;
}
 

下面考慮一道在區間中非單調函式

Strange fuction

Problem Description

Now, here is a fuction:   F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100) Can you find the minimum value when x is between 0 and 100.

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 only one real numbers Y.(0 < Y <1e10)

Output

Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.

Sample Input

2
100
200

Sample Output

-74.4291
-178.8534

題意：輸入一個Y，求 F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)的最小值，精度保留4位

思路：二分搜尋，這裡需注意的是函式在區間中並非單調，不能直接二分，需要我們來求其單調區間

          F(x) = 6*x7 + 8*x6 + 7*x3 + 5*x2 - y*x 一階導           F’(x) = 42*x6 + 48*x5 + 21*x2 + 10*x – y

         所以我們只需求出其F’(x) == 0的點就可得到最小值

#include <stdio.h>
#include <math.h>
const double eps = 1e-8;
double y;
double cal(double x){
    return 42.0*pow(x,6.0)+48.0*pow(x,5.0)+21.0*pow(x,2.0)+10.0*x;
}
double ans(double x){
    return 6.0*pow(x,7.0)+8.0*pow(x,6.0)+7.0*pow(x,3.0)+5.0*pow(x,2.0)-y*x;
}
int main(){
    int T;
    double f,l,mid;
    scanf("%d",&T);
    while(T--){
        scanf("%lf",&y);
        if(cal(100.0)-y<=0.0){  //表明在【0，100】內為遞減，可直接輸出
            printf("%.4lf\n",ans(100.0));
            continue;
        }
        f=0.0,l=100.0;
        while(l-f>eps){
            mid=(f+l)/2.0;
            if(cal(mid)-y<0.0){
                f=mid;
            }else{
                l=mid;
            }
        }
        printf("%.4lf\n",ans(mid));
    }
    return 0;
}