解题思路:给出一个方程 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,求方程的解。 首先判断方程是否有解,因为该函数在实数范围内是连续的,所以只需使y的值满足f(0)<=y<=f(100),就一定能找到该方程的解,否则就无解。 然后是求解过程, 假设一个区间[a,b],mid=(a+b)/2,如果f(a)*f(b)<0,那么函数f(x)在区间[a,b]至少存在一个零点,如果f(a)<0,说明0点在其右侧,那么将a的值更新为当前mid的值,如果f(a)>0,说明0点在其左侧,将b的值更新为mid的值。画出图像更好分析。
Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9490 Accepted Submission(s): 4382
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!
#include<stdio.h> #include<string.h> double f(double x) { return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6; } int main() { int ncase; double y,ans,left,right,mid; scanf("%d",&ncase); while(ncase--) { scanf("%lf",&y); ans=0; left=0; right=100; if(f(0)<=y&&y<=f(100)) { while(right-left>0.000000001) { { mid=(left+right)*0.5; ans=f(mid); if(ans-y<0) left=mid; else right=mid; } } printf("%.4lf\n",mid); } else printf("No solution!\n"); } }