[kuangbin带你飞]专题一 简单搜索 E - Find The Multiple poj 1426
2015-11-12 21:26
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E - Find The Multiple
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
1426
Appoint description:
System Crawler (2015-11-08)
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
Sample Output
题意
给出一个整数n。求出任意一个它的倍数m,要求m必须只由十进制的'0'或'1'组成。
一看 怎么也想不出来
再看才n<=200 果断 打表 AC
大神的
解题方法: BFS+同余模定理
/article/1968908.html
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
1426
Appoint description:
System Crawler (2015-11-08)
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
题意
给出一个整数n。求出任意一个它的倍数m,要求m必须只由十进制的'0'或'1'组成。
一看 怎么也想不出来
再看才n<=200 果断 打表 AC
大神的
解题方法: BFS+同余模定理
/article/1968908.html
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