#include
Int main ()
{
int i,j;
Double sum, c, a, b;
Lf the scanf (" % ", & amp; c);
Sum=1; B=1; A=1;
For (I=1, j=3; i<=100; I++, j +=2)
{
B *=I * 1.0;
A * * 1.0=j;
Sum=the sum a + b/a;
If (2.0 * sum> C)
{
Printf (" %. 7 lf % d ", 2.0 * (sum - b/a), I);
break;
}
}
printf("\n");
}
If the code behind the interpretation
CodePudding user response:
If the back is a printing, this can't explain, suggest pay attention to look at the requirements, see 2.0 * (sum - b/a) should be the contents of the ~ in the demandCodePudding user response:
I hope it can help you: https://blog.csdn.net/it_xiangqiang/category_10581430.htmlI hope it can help you: https://blog.csdn.net/it_xiangqiang/category_10768339.html