已知不等式 1!+2!+3!+...+m!‹n,请编程对用户指定的n值计算并输出满足该不等式的m的整数解。
#include <stdio.h>
int main()
{
long long int n, sum, total;
int i, j, count;
scanf("%lld", &n);
total = 0;
count = 0;
for (i = 1; total < n; i++)
{
for (j = i, sum=1; j >=1; j--)
sum *= j;
total += sum;
}
printf("m<=%d\n", i-2);
return 0;
}
注意数据溢出,这里要采用long long int或者double