C语言求1-1⼀2!+1⼀3!-1⼀4!...(-1)^(n-1)*1⼀n!,最后一个精度为0.000001

2025-04-08 16:11:19
推荐回答(1个)
回答1:

int i = 0,num=1,sign = 1;

double sum=0.0;

for(i=1;;i++)
{
if((1.0/num)<0.00000001)
break;
else
sum+=(1.0/(num*=i))*sign;

sign = -sign;
}