C语言 利用e=1+ (1/1！)+(1/2!) +....+(1/n!)直到最后一项绝对值小于10的

#include <stdio.h> #include <math.h> main() {int count = 1,n = 1;double e = 1.0,term = 1.0;while(fabs(term) >= 1e-5){term = term / n; //累加项e = e + term;//累加count++;//累加的次数n++;}printf("e = %f\ncount = %d\n",e,count); }