POJ 1426：Find The Multiple

Find The Multiple

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

Submit Status

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

题意是要找到input数的一个倍数，该倍数其十进制只含有0和1。

刚刚看完了广度优先搜索，A了几道题之后，做这个题卡住了，原因是不知道怎么用广度搜索表示十进制只有0 1的数。结果看了discuss才知道一开始用1，之后1导出10 11，10导出100 101，11导出110 111.这么一直下去就能有所有的数了。觉得嗯嗯，不错(。。。)

之后是自己建队列，结果莫名其妙地MLE，我也真是什么错误都见识过了，不知道怎么办，索性因为输入时大于等于1小于等于200的，得到结果打表吧。

打表代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <queue>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

long long temp;
queue<long long> q;

int main()
{
	freopen("i.txt","r",stdin);
	freopen("o.txt","w",stdout);

	__int64 test,temp;
	while(1)
	{
		cin>>test;

		if(test==0)
			break;
		while(q.size())q.pop();

		q.push(1);
		while(1)
		{
			temp=q.front();
			q.pop();
			if(temp%test==0)
			{
				cout<<"a["<<test<<"]="<<temp<<";"<<endl;
				break;
			}
			else
			{
				q.push(temp*10);//亮点在于所有0 1组成的十进制的数都可以这样广度搜索出来，这种想法很亮。
				q.push(temp*10+1);
			}

		}
	}
	return 0;
}

最终代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <queue>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

long long a[205];

int main()
{
	a[1]=1;
	a[2]=10;
	a[3]=111;
	a[4]=1100;
	a[5]=100;
	a[6]=11010;
	a[7]=11011;
	a[8]=11000;
	a[9]=111111111;
	a[10]=1000000;
	a[11]=1000010;
	a[12]=1001100;
	a[13]=1010100;
	a[14]=1011010;
	a[15]=1100010;
	a[16]=1110000;
	a[17]=1110100;
	a[18]=10111111110;
	a[19]=100111;
	a[20]=101000;
	a[21]=101010;
	a[22]=110000;
	a[23]=110101;
	a[24]=111000;
	a[25]=111100;
	a[26]=11001000010;
	a[27]=11010111111;
	a[28]=11011000000;
	a[29]=11011000111;
	a[30]=11011001010;
	a[31]=11011001011;
	a[32]=11011100000;
	a[33]=11011100001;
	a[34]=11011101110;
	a[35]=11011110110;
	a[36]=110111111100;
	a[37]=111000000000;
	a[38]=111000010010;
	a[39]=111001000110;
	a[40]=111001001000;
	a[41]=111001010110;
	a[42]=111010011000;
	a[43]=111101111110;
	a[44]=111110000100;
	a[45]=111110011110;
	a[46]=1001111110;
	a[47]=1010001001;
	a[48]=1010010000;
	a[49]=1010010001;
	a[50]=1010010100;
	a[51]=1010111100;
	a[52]=1011101000;
	a[53]=1100000001;
	a[54]=11011111110;
	a[55]=11100000010;
	a[56]=11100110000;
	a[57]=11110001100;
	a[58]=11110100100;
	a[59]=11111100011;
	a[60]=11111111100;
	a[61]=1000101100;
	a[62]=1010011000;
	a[63]=1111011111;
	a[64]=1010000000000;
	a[65]=1010000000100;
	a[66]=1010000101110;
	a[67]=1010001101100;
	a[68]=1010001110000;
	a[69]=1010011100001;
	a[70]=1010011110010;
	a[71]=1010110000110;
	a[72]=1110111111000;
	a[73]=1111010000110;
	a[74]=1111010100000;
	a[75]=1111010111100;
	a[76]=1111011111000;
	a[77]=1111100000010;
	a[78]=1111101010110;
	a[79]=1111101101100;
	a[80]=1111101110000;
	a[81]=1111110001101;
	a[82]=1111110111010;
	a[83]=10010111001;
	a[84]=10011101100;
	a[85]=10011110110;
	a[86]=10100101010;
	a[87]=10100111010;
	a[88]=10110001000;
	a[89]=10111111110;
	a[90]=11011111110;
	a[91]=11011111111;
	a[92]=11101111000;
	a[93]=11111010111;
	a[94]=1000011010;
	a[95]=1000101100;
	a[96]=1001100000;
	a[97]=1001101110;
	a[98]=1010110010;
	a[99]=1101111111111111111;
	a[100]=1000000;
	a[101]=1000001;
	a[102]=1000110;
	a[103]=1110000000010110110;
	a[104]=1110000000011110000;
	a[105]=1110000000011110110;
	a[106]=1110000001010000100;
	a[107]=1110000001011001101;
	a[108]=1110000001011101100;
	a[109]=1110000001110010101;
	a[110]=1110000001110011000;
	a[111]=1110000001110011100;
	a[112]=1110000010000000000;
	a[113]=1110000010001000011;
	a[114]=1110000010001101110;
	a[115]=1110000010010110110;
	a[116]=1110000010011101100;
	a[117]=1110000010011110010;
	a[118]=1110000010101000110;
	a[119]=1110000010101100111;
	a[120]=1110000010101111000;
	a[121]=1110000010110100010;
	a[122]=1110000010111000110;
	a[123]=1110000010111001100;
	a[124]=1110000010111111100;
	a[125]=1110000011000000000;
	a[126]=1110000011000011110;
	a[127]=1110000011000101110;
	a[128]=1110000011010000000;
	a[129]=1110000011101001001;
	a[130]=1110000011101111000;
	a[131]=1110000011110111100;
	a[132]=1110000100000001100;
	a[133]=1110000100000111110;
	a[134]=1110000100111100000;
	a[135]=1110000100111100010;
	a[136]=1110000101001100000;
	a[137]=1110000101100000111;
	a[138]=1110000101110100100;
	a[139]=1110000101110100101;
	a[140]=1110000101110110000;
	a[141]=1110000110001100101;
	a[142]=1110000111000100100;
	a[143]=1110000111110000110;
	a[144]=1110000111110010000;
	a[145]=1110000111111011010;
	a[146]=1110001000000010110;
	a[147]=1110001000010110110;
	a[148]=1110001000010111100;
	a[149]=1110001000011010010;
	a[150]=1110001000011011100;
	a[151]=1110001000011100010;
	a[152]=1110001000101101000;
	a[153]=1110001000101101001;
	a[154]=1110001001000001110;
	a[155]=1110001001001001100;
	a[156]=1110001001011110000;
	a[157]=1110001010010000110;
	a[158]=1110001010011010100;
	a[159]=1110001010101010001;
	a[160]=1110001010101100000;
	a[161]=1110001011010110100;
	a[162]=1110001011011001000;
	a[163]=1110001011101010101;
	a[164]=1110001011101110000;
	a[165]=1110001011110101110;
	a[166]=1110001100000111000;
	a[167]=1110001100100001001;
	a[168]=1110001100101110000;
	a[169]=1110001101001001001;
	a[170]=1110001101001110000;
	a[171]=1110001101010000110;
	a[172]=1110001101111101000;
	a[173]=1110001101111110010;
	a[174]=1110001110011000100;
	a[175]=1110001110011010000;
	a[176]=1110001111000100000;
	a[177]=1110001111010110011;
	a[178]=1110010001011110100;
	a[179]=1110010010110111011;
	a[180]=1110010010111000100;
	a[181]=1110010011001101001;
	a[182]=1110010011010000100;
	a[183]=1110010011011010000;
	a[184]=1110010011011110000;
	a[185]=1110010011100001100;
	a[186]=1110010011111011100;
	a[187]=1110010100000110001;
	a[188]=1110010101001111000;
	a[189]=1110010110000101100;
	a[190]=1110010110000110100;
	a[191]=1110010110010010001;
	a[192]=1110010110111000000;
	a[193]=1110011000000010110;
	a[194]=1110011000001000010;
	a[195]=1110011000001111000;
	a[196]=1110011000010010000;
	a[197]=1101000101;
	a[198]=1111111111111111110;
	a[199]=111000011;
	a[200]=1000;
	long long test;

	while(1)
	{
		cin>>test;

		if(test==0)
			break;
		cout<<a[test]<<endl;
	}

	return 0;
}

还是觉得0 1表示那里是亮点，乘以10，乘以10+1就能推出所有了，我当时怎么就没想到呢？？？

版权声明：本文为博主原创文章，未经博主允许不得转载。