π/4=1-1/3+1/5-1/7+一直加到最后一个数的绝对值小于10^-6 有什么错误啊#include#includeint main(){int n;\x05double s,b=1.0,a,sum;sum=0;\x05for(n=2;fabs(b)>=1e-6;n++)\x05{\x05\x05sum=sum+b;\x05\x05a=2*n-1;\x05\x05b=-1/a;\x05}\x05s=sum
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![π/4=1-1/3+1/5-1/7+一直加到最后一个数的绝对值小于10^-6 有什么错误啊#include#includeint main(){int n;\x05double s,b=1.0,a,sum;sum=0;\x05for(n=2;fabs(b)>=1e-6;n++)\x05{\x05\x05sum=sum+b;\x05\x05a=2*n-1;\x05\x05b=-1/a;\x05}\x05s=sum](/uploads/image/z/11403599-23-9.jpg?t=%CF%80%2F4%3D1-1%2F3%2B1%2F5-1%2F7%2B%E4%B8%80%E7%9B%B4%E5%8A%A0%E5%88%B0%E6%9C%80%E5%90%8E%E4%B8%80%E4%B8%AA%E6%95%B0%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%E5%B0%8F%E4%BA%8E10%5E-6+%E6%9C%89%E4%BB%80%E4%B9%88%E9%94%99%E8%AF%AF%E5%95%8A%23include%23includeint+main%28%29%7Bint+n%3B%5Cx05double+s%2Cb%3D1.0%2Ca%2Csum%3Bsum%3D0%3B%5Cx05for%28n%3D2%3Bfabs%28b%29%3E%3D1e-6%3Bn%2B%2B%29%5Cx05%7B%5Cx05%5Cx05sum%3Dsum%2Bb%3B%5Cx05%5Cx05a%3D2%2An-1%3B%5Cx05%5Cx05b%3D-1%2Fa%3B%5Cx05%7D%5Cx05s%3Dsum)
π/4=1-1/3+1/5-1/7+一直加到最后一个数的绝对值小于10^-6 有什么错误啊#include#includeint main(){int n;\x05double s,b=1.0,a,sum;sum=0;\x05for(n=2;fabs(b)>=1e-6;n++)\x05{\x05\x05sum=sum+b;\x05\x05a=2*n-1;\x05\x05b=-1/a;\x05}\x05s=sum
π/4=1-1/3+1/5-1/7+一直加到最后一个数的绝对值小于10^-6 有什么错误啊
#include
#include
int main()
{int n;
\x05double s,b=1.0,a,sum;
sum=0;
\x05for(n=2;fabs(b)>=1e-6;n++)
\x05{\x05
\x05sum=sum+b;
\x05
\x05a=2*n-1;
\x05
\x05b=-1/a;
\x05}
\x05s=sum*4;
\x05printf("%10.8f\n",s);
\x05return 0;
}
π/4=1-1/3+1/5-1/7+一直加到最后一个数的绝对值小于10^-6 有什么错误啊#include#includeint main(){int n;\x05double s,b=1.0,a,sum;sum=0;\x05for(n=2;fabs(b)>=1e-6;n++)\x05{\x05\x05sum=sum+b;\x05\x05a=2*n-1;\x05\x05b=-1/a;\x05}\x05s=sum
b=-1/a; 这里错了,因为a是正数,所以b也是正数,题目要求是1正1负1交替
所以应该改为
if(n%2==0) b=-1/a;
else b=1/a;