C语言循环求和SUM=1+(1/2)^2+(1/3)^3+(1/4)^4+(1/5)^5……一直算到精确度为0.000001为止/*SUM=1+(1/2)^2+(1/3)^3+(1/4)^4+(1/5)^5……*/#include#include#define ACCURARY 0.000001main(){\x09int i=1,j;\x09double SUM=0,term=1;\x09for(i=1;term>
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![C语言循环求和SUM=1+(1/2)^2+(1/3)^3+(1/4)^4+(1/5)^5……一直算到精确度为0.000001为止/*SUM=1+(1/2)^2+(1/3)^3+(1/4)^4+(1/5)^5……*/#include#include#define ACCURARY 0.000001main(){\x09int i=1,j;\x09double SUM=0,term=1;\x09for(i=1;term>](/uploads/image/z/5169107-11-7.jpg?t=C%E8%AF%AD%E8%A8%80%E5%BE%AA%E7%8E%AF%E6%B1%82%E5%92%8CSUM%3D1%2B%281%2F2%29%5E2%2B%281%2F3%29%5E3%2B%281%2F4%29%5E4%2B%281%2F5%29%5E5%E2%80%A6%E2%80%A6%E4%B8%80%E7%9B%B4%E7%AE%97%E5%88%B0%E7%B2%BE%E7%A1%AE%E5%BA%A6%E4%B8%BA0.000001%E4%B8%BA%E6%AD%A2%2F%2ASUM%3D1%2B%281%2F2%29%5E2%2B%281%2F3%29%5E3%2B%281%2F4%29%5E4%2B%281%2F5%29%5E5%E2%80%A6%E2%80%A6%2A%2F%23include%23include%23define+ACCURARY+0.000001main%28%29%7B%5Cx09int+i%3D1%2Cj%3B%5Cx09double+SUM%3D0%2Cterm%3D1%3B%5Cx09for%28i%3D1%3Bterm%3E)
C语言循环求和SUM=1+(1/2)^2+(1/3)^3+(1/4)^4+(1/5)^5……一直算到精确度为0.000001为止/*SUM=1+(1/2)^2+(1/3)^3+(1/4)^4+(1/5)^5……*/#include#include#define ACCURARY 0.000001main(){\x09int i=1,j;\x09double SUM=0,term=1;\x09for(i=1;term>
C语言循环求和
SUM=1+(1/2)^2+(1/3)^3+(1/4)^4+(1/5)^5……一直算到精确度为0.000001为止
/*SUM=1+(1/2)^2+(1/3)^3+(1/4)^4+(1/5)^5……*/
#include
#include
#define ACCURARY 0.000001
main()
{
\x09int i=1,j;
\x09double SUM=0,term=1;
\x09for(i=1;term>=ACCURARY;i++)
\x09{
\x09\x09for(j=1;j
C语言循环求和SUM=1+(1/2)^2+(1/3)^3+(1/4)^4+(1/5)^5……一直算到精确度为0.000001为止/*SUM=1+(1/2)^2+(1/3)^3+(1/4)^4+(1/5)^5……*/#include#include#define ACCURARY 0.000001main(){\x09int i=1,j;\x09double SUM=0,term=1;\x09for(i=1;term>
你用1/i 是不是直接转换成int型了啊 你是试试1.0/i~~这样应该是默认转成double型的了
另外最后printf("SUM=%-6f\n",SUM); 是输入小数点后6位