poj1426

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 10685 Accepted: 4404 Special Judge

Description

Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.

Input

The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.

Output

For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.

Sample Input

2
6
19
0

Sample Output

10
100100100100100100
111111111111111111
没有仔细分析题目,我一看最后的output不会超过100位,我就在想怎么存呢
想了好久也没想出好方法,如果数组存的话,可能会超内存,判断也可能会超时,最后现不到好办法
我还想要不要用double存,然后只是判断麻烦一些而以,
结果去网上找题解发现,只要longlong就能出正解
呃……我是沙茶
还有队列开了好大,要定义成全局变量
 1#include<stdio.h>
 2#include<string.h>
 3#include<math.h>
 4int n;
 5long long now,q[1000000];
 6void bfs()
 7{
 8    int head,tail;
 9    head=0;
10    tail=1;
11    q[tail]=1;
12    while(head<tail)
13    {
14        head++;
15        now=q[head];
16        now=now*10;
17        if(now%n==0)
18        {
19            break;
20        }

21        tail++;
22        q[tail]=now;
23        tail++;
24        q[tail]=now+1;
25    }

26    printf("%I64d\n",now);
27}

28int main()
29{
30    while(scanf("%d",&n)!=EOF&&n!=0)
31    {
32        bfs();
33    }

34    return 0;
35}

36

posted on 2012-03-15 19:05 jh818012 阅读(627) 评论(2)  编辑 收藏 引用

评论

# re: poj1426 2012-04-02 14:08 王私江

17 if(now%n==0)
18 {
19 break;
20 }
你这里判断明显慢了很多啊,在下面判断now*10和now*10+1这样要快很多。  回复  更多评论   

# re: poj1426 2012-09-20 20:13 season

我嚓,,辉哥,,居然搜到你的题解了  回复  更多评论   


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