编程计算:1 - 1/2 + 1/3 -1/4 + ...+ 1/99 - 1/100 +...,直到最后一项的绝对值小于10-4为止.哪里错了啊.#include #include int main(){float m=1.0,sum=0;int n=1;while(fabs(m/n)>=1e-4){m=pow(-1,n+1);sum=sum+m/n;n++;}printf("sum=%f\n",s
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![编程计算:1 - 1/2 + 1/3 -1/4 + ...+ 1/99 - 1/100 +...,直到最后一项的绝对值小于10-4为止.哪里错了啊.#include #include int main(){float m=1.0,sum=0;int n=1;while(fabs(m/n)>=1e-4){m=pow(-1,n+1);sum=sum+m/n;n++;}printf(](/uploads/image/z/135114-42-4.jpg?t=%E7%BC%96%E7%A8%8B%E8%AE%A1%E7%AE%97%EF%BC%9A1+-+1%2F2+%2B+1%2F3+-1%2F4+%2B+...%2B+1%2F99+-+1%2F100+%2B...%2C%E7%9B%B4%E5%88%B0%E6%9C%80%E5%90%8E%E4%B8%80%E9%A1%B9%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%E5%B0%8F%E4%BA%8E10-4%E4%B8%BA%E6%AD%A2.%E5%93%AA%E9%87%8C%E9%94%99%E4%BA%86%E5%95%8A.%23include+%23include+int+main%28%29%7Bfloat+m%3D1.0%2Csum%3D0%3Bint+n%3D1%3Bwhile%28fabs%28m%2Fn%29%3E%3D1e-4%29%7Bm%3Dpow%28-1%2Cn%2B1%29%3Bsum%3Dsum%2Bm%2Fn%3Bn%2B%2B%3B%7Dprintf%28%22sum%3D%25f%5Cn%22%2Cs)
编程计算:1 - 1/2 + 1/3 -1/4 + ...+ 1/99 - 1/100 +...,直到最后一项的绝对值小于10-4为止.哪里错了啊.#include #include int main(){float m=1.0,sum=0;int n=1;while(fabs(m/n)>=1e-4){m=pow(-1,n+1);sum=sum+m/n;n++;}printf("sum=%f\n",s
编程计算:1 - 1/2 + 1/3 -1/4 + ...+ 1/99 - 1/100 +...,直到最后一项的绝对值小于10-4为止.
哪里错了啊.
#include
#include
int main()
{
float m=1.0,sum=0;
int n=1;
while(fabs(m/n)>=1e-4)
{
m=pow(-1,n+1);
sum=sum+m/n;
n++;
}
printf("sum=%f\n",sum);
return 0;
}
编程计算:1 - 1/2 + 1/3 -1/4 + ...+ 1/99 - 1/100 +...,直到最后一项的绝对值小于10-4为止.哪里错了啊.#include #include int main(){float m=1.0,sum=0;int n=1;while(fabs(m/n)>=1e-4){m=pow(-1,n+1);sum=sum+m/n;n++;}printf("sum=%f\n",s
#include
#include
int main()
{
double m=1,sum=0;
int n=1;
while(fabs(m/n)>=1e-4)
{
m=pow(-1,n+1);
sum=sum+m/n;
n++;
}
printf("sum=%f\n",sum);
return 0;
}
一点分都不给我 哼哼 ~~~~~~
记住数学函数返回值是double 型 就好啦!这个是不是应该要求精确度 啊
就当交个朋友