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 0.000001main(){\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