Can you solve this equation?
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.
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!
题目大意:给出方程y = 8 * x ^ 4 + 7 * x ^ 3 + 2 * x ^ 2 + 3 * x + 6,其中fabs(y) <= 1e10,求x在区间[0,100]内的解,输出4位小数。
解题思路:在区间[0,100]内二分x,将x带入方程,判断结果与y的大小关系,若ans > y,则ub = mid;若ans < y,则lb = mid。由于fabs(y) <= 1e10,所以经过40次二分即可带到精度要求。
代码如下:
#include <cstdio>
#include <cmath>
double y;
double fun(double x)
{
return 8 * pow(x,4.0) + 7 * pow(x,3.0) + 2 * pow(x,2.0) + 3 * x + 6;
}
int main()
{
double lb,ub;
int t;
scanf("%d",&t);
while(t--){
scanf("%lf",&y);
if(fun(0) <= y && y <= fun(100)){
lb = 0,ub = 100;
for(int i = 0;i < 40;i++){
double mid = (ub + lb) / 2.0;
if(fun(mid) > y)
ub = mid ;
else
lb = mid ;
}
printf("%.4lf\n",lb);
}else{
printf("No solution!\n");
}
}
return 0;
}