HDU2199:Can you solve this equation?(二分)
2013-11-24 14:36
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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
Sample Output
一开始跑去百度一元四次方程的解,直接就被吓尿了,然后知道用二分做之后,发现二分还能解决这样的题目,数学无法做到的事情用程序可以很简单的就做到了
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!
一开始跑去百度一元四次方程的解,直接就被吓尿了,然后知道用二分做之后,发现二分还能解决这样的题目,数学无法做到的事情用程序可以很简单的就做到了
#include <stdio.h> #include <math.h> #include <algorithm> #include <string.h> #include <math.h> using namespace std; double cal(double x) { return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6; } int main() { int t; scanf("%d",&t); while(t--) { double n; scanf("%lf",&n); if(cal(0)>n || cal(100)<n) { printf("No solution!\n"); continue; } double l = 0.0,r = 100.0; double mid = (l+r)/2; while(fabs(cal(mid)-n)>1e-5) { if(cal(mid)>n) r = mid-1; else l = mid+1; mid = (l+r)/2; } printf("%.4f\n",mid); } return 0; }
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