求π²/6的值= 1+(½)²+(1/3)²+(1/4)²+.+(1/200)²c++代码 为什么出错#include "stdafx.h"#include <iostream.h>int powers (int n,int k){\x05long m=1,z;\x05for (int i=1;i<=k;i++)\x05\x05m=n*m;\x05\x05z
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![求π²/6的值= 1+(½)²+(1/3)²+(1/4)²+.+(1/200)²c++代码 为什么出错#include "stdafx.h"#include <iostream.h>int powers (int n,int k){\x05long m=1,z;\x05for (int i=1;i<=k;i++)\x05\x05m=n*m;\x05\x05z](/uploads/image/z/204352-16-2.jpg?t=%E6%B1%82%CF%80%26%23178%3B%2F6%E7%9A%84%E5%80%BC%3D+1%2B%28%26%23189%3B%29%26%23178%3B%2B%281%2F3%29%26%23178%3B%2B%281%2F4%29%26%23178%3B%2B.%2B%281%2F200%29%26%23178%3Bc%2B%2B%E4%BB%A3%E7%A0%81+%E4%B8%BA%E4%BB%80%E4%B9%88%E5%87%BA%E9%94%99%23include+%26quot%3Bstdafx.h%26quot%3B%23include+%26lt%3Biostream.h%26gt%3Bint+powers+%28int+n%2Cint+k%29%7B%5Cx05long+m%3D1%2Cz%3B%5Cx05for+%28int+i%3D1%3Bi%26lt%3B%3Dk%3Bi%2B%2B%29%5Cx05%5Cx05m%3Dn%2Am%3B%5Cx05%5Cx05z)
求π²/6的值= 1+(½)²+(1/3)²+(1/4)²+.+(1/200)²c++代码 为什么出错#include "stdafx.h"#include <iostream.h>int powers (int n,int k){\x05long m=1,z;\x05for (int i=1;i<=k;i++)\x05\x05m=n*m;\x05\x05z
求π²/6的值= 1+(½)²+(1/3)²+(1/4)²+.+(1/200)²c++代码 为什么出错
#include "stdafx.h"
#include <iostream.h>
int powers (int n,int k)
{\x05long m=1,z;
\x05for (int i=1;i<=k;i++)
\x05\x05m=n*m;
\x05\x05z=1/m;
\x05return z;
}
double sum (int n,int k)
{\x05
\x05double sum=0;
\x05for(int i=1;i<=n;i++)
\x05\x05sum=sum+powers(i,k);
\x05return sum;
}
int main(int argc,char* argv[])
{
\x05int n=200,k=2 ;
\x05\x05cout<<"π2/6的值等于"<<sum(n,k)<<endl;
\x05return 0;
}
错在哪里?
这是运行结果
求π²/6的值= 1+(½)²+(1/3)²+(1/4)²+.+(1/200)²c++代码 为什么出错#include "stdafx.h"#include <iostream.h>int powers (int n,int k){\x05long m=1,z;\x05for (int i=1;i<=k;i++)\x05\x05m=n*m;\x05\x05z
double powers (int n,int k)
{
double m=1.0,z;
for (int i=1; i<=k; i++)
m=n*m;
z=1.0/m;
return z;
}
要用浮点数来表示.如果用整型表示1/m:1/1==1,1/2==0,1/3==0,.,