OJ1058: 求解不等式

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

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转载自blog.csdn.net/weixin_46272577/article/details/110530760