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hdu2117 Just a Numble

2014年09月05日 ⁄ 综合 ⁄ 共 834字 ⁄ 字号小中大 ⁄ 评论关闭

Just a Numble

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2077 Accepted Submission(s): 970

Problem Description

Now give you two integers n m, you just tell me the m-th number after radix point in 1/n,for example n=4,the first numble after point is 2,the second is 5,and all 0 followed

Input

Each line of input will contain a pair of integers for n and m(1<=n<=10^7,1<=m<=10^5)

Output

For each line of input, your program should print a numble on a line,according to the above rules

Sample Input

4 2
5 7
123 123

Sample Output

5
0
8

模拟除法，刚开始还以为很麻烦呢，还找了循环节，循环节前有几位不循环，还有不是循环小数的，还有1。。。结果学长说一直算下去就行！！！

#include<stdio.h>
#include<string.h>
int s[100004];
int main()
{
	int n,m,a,i;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(n==1)
		{
			printf("0\n");
			continue;
		}
		memset(s,0,sizeof(s));
		a=1;
		for(i=1;i<=m;i++)
		{
			a*=10;
			while(a<n)
			{
				a*=10;
				s[i++]=0;
			}
			s[i]=a/n;
			a%=n;
			if(!a) break;
		}
		printf("%d\n",s[m]);
	}
	return 0;
}