习题6.11

//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);}

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