#include#includevoid main(){double y,x,n;n=1;printf("Enter x
来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/10/06 11:06:05
#include
#include
void main()
{
double y,x,n;
n=1;
printf("Enter x(x>1)\n");
scanf("%f",&x);
for(y=1;y>=0.000001;n++)
{ y=y+(1/pow(x,n));
printf("%f\n",y);}
}
#include
void main()
{
double y,x,n;
n=1;
printf("Enter x(x>1)\n");
scanf("%f",&x);
for(y=1;y>=0.000001;n++)
{ y=y+(1/pow(x,n));
printf("%f\n",y);}
}
// 程序改了,看后面的注释
#include<stdio.h>
#include<math.h>
void main()
{
double y,x,n;
n=1;
printf("Enter x(x>1)\n");
scanf("%lf",&x); // 你上面定义的是 double 类型,函数内要改成 %lf
for(y=1;y>=0.000001;n++)
{ y=y+(1/pow(x,n));
printf("%lf\n",y);} // 你上面定义的是 double 类型,函数内要改成 %lf
}
#include<stdio.h>
#include<math.h>
void main()
{
double y,x,n;
n=1;
printf("Enter x(x>1)\n");
scanf("%lf",&x); // 你上面定义的是 double 类型,函数内要改成 %lf
for(y=1;y>=0.000001;n++)
{ y=y+(1/pow(x,n));
printf("%lf\n",y);} // 你上面定义的是 double 类型,函数内要改成 %lf
}
# include # include int main() { int n; printf("enter n:\n")
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下列程序输出结果是:#includevoid main(){char x=040;printf("%o\n",x
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