1 solutions
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0
C++ :
#include <iostream> #include <cmath> using namespace std; int i,j,n; double f(double x) { return(pow(2,x)+pow(3,x)-pow(4,x)); } main() { double x,r,l; r=2;l=1; x=(r+l)/2; while (f(x)!=0 && r-l>0.000000001) { if (f(r)*f(x)<0) l=x; else r=x; x=(r+l)/2; } printf("%0.10lf",x); }
- 1
Information
- ID
- 18963
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
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