//已知不等式 1!+2!+3!+...+m!<n,请编程对用户指定的n值计算并输出满足不等式m的值
#include "stdio.h"
void main()
{
int i,n,term=1,sum=0;
printf("Please Input n: \n");
scanf("%d",&n);
for (i=1;sum<n;i++)
{
term*=i;
sum+=term;
if(sum>=n)break;
printf("sum=%ld\n",sum);
}
printf("\n");
printf("m<=%d\n",i);
}
运行结果: