通过公式输出π的近似值,并输出#include "stdio.h"#include "math.h"int main(){\x05int n;\x05double sum=0;\x05\x05for(n=1;fabs(sqrt(6*sum)-3.1415926)>1E-6;n++)//∑(n=1--∞)1/n*n=π*π/6\x05\x05{\x05sum=sum+1.0/(n*n);}\x05\x05printf("%lf",s
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/20 14:56:11
![通过公式输出π的近似值,并输出#include](/uploads/image/z/8437738-58-8.jpg?t=%E9%80%9A%E8%BF%87%E5%85%AC%E5%BC%8F%E8%BE%93%E5%87%BA%CF%80%E7%9A%84%E8%BF%91%E4%BC%BC%E5%80%BC%2C%E5%B9%B6%E8%BE%93%E5%87%BA%23include+%22stdio.h%22%23include+%22math.h%22int+main%28%29%7B%5Cx05int+n%3B%5Cx05double+sum%3D0%3B%5Cx05%5Cx05for%28n%3D1%3Bfabs%28sqrt%286%2Asum%29-3.1415926%29%3E1E-6%3Bn%2B%2B%29%2F%2F%E2%88%91%28n%3D1--%E2%88%9E%291%2Fn%2An%3D%CF%80%2A%CF%80%2F6%5Cx05%5Cx05%7B%5Cx05sum%3Dsum%2B1.0%2F%28n%2An%29%3B%7D%5Cx05%5Cx05printf%28%22%25lf%22%2Cs)
通过公式输出π的近似值,并输出#include "stdio.h"#include "math.h"int main(){\x05int n;\x05double sum=0;\x05\x05for(n=1;fabs(sqrt(6*sum)-3.1415926)>1E-6;n++)//∑(n=1--∞)1/n*n=π*π/6\x05\x05{\x05sum=sum+1.0/(n*n);}\x05\x05printf("%lf",s
通过公式输出π的近似值,并输出
#include "stdio.h"
#include "math.h"
int main()
{
\x05int n;
\x05double sum=0;
\x05\x05for(n=1;fabs(sqrt(6*sum)-3.1415926)>1E-6;n++)//∑(n=1--∞)1/n*n=π*π/6
\x05\x05{\x05sum=sum+1.0/(n*n);}
\x05\x05printf("%lf",sqrt(6*sum));
\x05\x05return 0;
}
通过公式输出π的近似值,并输出#include "stdio.h"#include "math.h"int main(){\x05int n;\x05double sum=0;\x05\x05for(n=1;fabs(sqrt(6*sum)-3.1415926)>1E-6;n++)//∑(n=1--∞)1/n*n=π*π/6\x05\x05{\x05sum=sum+1.0/(n*n);}\x05\x05printf("%lf",s
//就是类型转换的问题
#include "stdio.h"
#include "math.h"
int main()
{
\x05int n;
\x05double sum=0.0;
\x05\x05for(n=1;fabs(sqrt(6*sum)-3.1415926)>1E-6;n++)//∑(n=1--∞)1/n*n=π*π/6
\x05\x05{\x05sum=sum+1.0/((double)n*(double)n);} //事先就要把n的值转换为double
\x05\x05printf("%lf",sqrt(6*sum));
\x05\x05return 0;
}
结果
3.141592请按任意键继续...