POJ - 1426 Find The Multiple

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本文介绍了一个编程问题FindTheMultiple的解决方案，该问题要求找到一个由0和1组成的正整数，使其能够被给定的不超过200的整数整除。使用递归深度优先搜索的方法来寻找符合条件的数字。

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Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 32518		Accepted: 13601		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

Sample Output

10
100100100100100100
111111111111111111

大意：给出一个整数不超过200，求它任意一个只由1或0组成的数字（能整除给出的整数）

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
int n,flag;
void dfs(unsigned __int64 x,int k)
{
	if(flag) return ;
	if(x%n==0)
	{
		printf("%I64u\n",x);
		flag=1;
		return ;
	 } 
	 if(k==19) return;// //为防止超出范围，循环深度应小于20
	 dfs(x*10,k+1);// //当前数字有两种选择方案，即下一个数选1或选0
	 dfs(x*10+1,k+1);
}
int main()
{
	while(~scanf("%d",&n))
	{
		if(n==0) break;
		flag=0;
		dfs(1,0);	
	}
	return 0;
 }

原文链接: http://blog.youkuaiyun.com/lmyclever/article/details/6744906

有符号型64位整数，值域为：-9223372036854775808 .. 9223372036854775807。

语言	GNU C/C++	Pascal	Visual C/C++
类型名称	__int64 or long long	int64	__int64
输入方法	scanf("%I64d", &x); or cin >> x;	read(x);	scanf("%I64d", &x);
输出方法	printf("%I64d", x); cout << x;	write(x);	printf("%I64d", x);

无符号型64位整数，值域为：0 .. 18446744073709551615。

语言	GNU C/C++	Pascal	Visual C/C++
类型名称	unsigned __int64 or unsigned long long	qword	unsigned __int64
输入方法	scanf("%I64u", &x); or cin >> x;	read(x);	scanf("%I64u", &x);
输出方法	printf("%I64u", x); or cout << x;	write(x);	printf("%I64u", x);