[kuangbin带你飞]专题一简单搜索E - Find The Multiple(POJ 1426)

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本文介绍了一种算法问题，即给定一个正整数n，如何找到一个仅由1和0组成的n的非零倍数m。文中提供了一个具体的实现方案，并通过预计算的方式生成了一个包含所有可能解的数据表。

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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:

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

题意：输出n的倍数且这个倍数只含1和0。

最高位肯定是1

方法一：直接枚举后面的位数1或者0。但是会超时我这里用的打表

打表

///打表
#include
  
   
#include
   
    
#include
    
     
using namespace std;

queue
     
      que;
int n;
int judge(string tmp)
{
    int ans = 0;
    int len = tmp.length();
    for(int i = 0; i < len; i++)
    {
        ans = (ans*10+tmp[i]-'0')%n;
    }
    if(ans) return 0;
    return 1;
}

int bfs()
{
    while(!que.empty()) que.pop();
    que.push("1");

    while(!que.empty())
    {
        string tmp = que.front(); que.pop();
        if(!judge(tmp))
        {
            que.push(tmp+"1");
            que.push(tmp+"0");
        }
        else
        {
            cout  << tmp << ',';
            return 0;
        }
    }
    return 0;
}

int main(void)
{
    int i, j;
    freopen("in.txt", "w", stdout);
    for(n = 1; n <= 200; n++)
//    while(scanf("%d", &n), n)
    {
        if(n == 1) {printf("1,");continue;}
        bfs();
    }
    return 0;
}

#include
   
    
#include
    
     
using namespace std;

long long str[222] = {0, 1,10,111,100,10,1110,1001,1000,111111111,10,11,11100,1001,10010,1110,10000,11101,1111111110,11001,100,10101,110,110101,111000,100,10010,1111111101,100100,1101101,1110,111011,100000,111111,111010,10010,11111111100,111,110010,10101,1000,11111,101010,1101101,1100,1111111110,1101010,10011,1110000,1100001,100,100011,100100,100011,11111111010,110,1001000,11001,11011010,11011111,11100,100101,1110110,1111011111,1000000,10010,1111110,1101011,1110100,11111001,10010,10011,111111111000,10001,1110,11100,1100100,1001,101010,11101001,10000,1111111101,111110,101011,1010100,111010,11011010,11010111,11000,11010101,1111111110,1001,11010100,10000011,100110,110010,11100000,11100001,11000010,111111111111111111,100,101,1000110,11100001,1001000,101010,1000110,110100111,111111110100,1111111011,110,111,10010000,1011011,110010,1101010,110110100,11111110101,110111110,110001101,111000,11011,1001010,10011100011,11101100,1000,11110111110,11010011,10000000,100100001,10010,101001,11111100,11101111,11010110,11111111010,11101000,10001,111110010,110110101,100100,10011,100110,1001,1111111110000,11011010,100010,1100001,11100,110111,11100,1110001,11001000,11111011011,10010,1110110,1010100,10101101011,111010010,100011,100000,11101111,11111111010,1010111,1111100,1111110,1010110,11111011,10101000,10111101,111010,1111011111,110110100,1111111011,110101110,100100,110000,101110011,110101010,11010111,11111111100,1001111,10010,100101,110101000,1110,100000110,1001011,1001100,1111111100001,110010,11101111,111000000,11001,111000010,101010,110000100,1101000101,1111111111111111110,111000011,1000};

int main(void)
{
    int n;
    while(cin >> n && n)
    {
        cout << str[n] << endl;
    }
    return 0;
}