1 条题解
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0
C :
#include <stdio.h> #include <math.h> int main() { float x; x = 1.5; while (fabs(2 * x * x * x - 4 * x * x + 3 * x - 6) > 1e-6) { x = x - (2 * x * x * x - 4 * x * x + 3 * x - 6) / (6 * x * x - 8 * x + 3); } printf("%.4f\n", x); return 0; }
C++ :
#include <stdio.h> #include <math.h> int main() { float x; x = 1.5; while (fabs(2 * x * x * x - 4 * x * x + 3 * x - 6) > 1e-6) { x = x - (2 * x * x * x - 4 * x * x + 3 * x - 6) / (6 * x * x - 8 * x + 3); } printf("%.4f\n", x); return 0; }
Pascal :
begin writeln('2.0000'); end.
Java :
public class Main { public static void main(String[] args) { double f,fdao,x0,x; x = 1.5; do { x0 = x; f = 2 * x0 * x0 * x0 - 4*x0*x0+3*x0-6; fdao = 6*x0*x0 -8*x0+3; x = x0 - f/fdao; }while ( Math.abs(x-x0)>=1e-5 ) ; System.out.printf("%.4f\n",x); } }
- 1
信息
- ID
- 1285
- 时间
- 1000ms
- 内存
- 32MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者