Find The Multiple
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 43 Accepted Submission(s) : 21
Special Judge
Problem 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
Source
PKU
题意：构造一个十进制由0和1组成的整数m，让m能够被n整除；题目中给出了m是小于等于100位的。
思路：bfs可以做出的，只是100位这个条件让我很头疼，我先用string试了试交了之后是超时，后来看了看答案用long long也给过了，所以m肯定没有100位的；
string类型代码：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<queue> 5 6 using namespace std; 7 8 queue<string>s; 9 int a,c; 10 string b; 11 12 13 int chu(string s) 14 { 15 c=0; 16 for(int i=0;i<s.length();i++){ 17 c*=10; 18 c+=s[i]-'0'; 19 c%=a; 20 } 21 if(c==0) 22 return 1; 23 else 24 return 0; 25 } 26 27 28 int bfs() 29 { 30 while(!s.empty()){ 31 b=s.front(); 32 s.pop(); 33 if(b.length()>100) 34 continue; 35 if(chu(b+'1')==1){ 36 cout<<b+'1'<<endl; 37 while(!s.empty()){ 38 s.pop(); 39 } 40 return 0; 41 } 42 else{ 43 s.push(b+'1'); 44 } 45 46 if(chu(b+'0')==1){ 47 cout<<b+'0'<<endl; 48 while(!s.empty()){ 49 s.pop(); 50 } 51 return 0; 52 } 53 else{ 54 s.push(b+'0'); 55 } 56 } 57 return 0; 58 } 59 60 61 int main() 62 { 63 // freopen("input.txt","r",stdin); 64 while(cin>>a){ 65 if(a==0) 66 break; 67 else{ 68 s.push("1"); 69 bfs(); 70 } 71 } 72 return 0; 73 }
long long类型代码：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<queue> 5 6 using namespace std; 7 8 queue<long long>s; 9 int a,c; 10 long long b; 11 12 13 14 int bfs() 15 { 16 while(!s.empty()){ 17 b=s.front(); 18 s.pop(); 19 if(b%a==0){ 20 cout<<b<<endl; 21 while(!s.empty()){ 22 s.pop(); 23 } 24 return 0; 25 } 26 else{ 27 s.push(b*10); 28 s.push(b*10+1); 29 } 30 } 31 return 0; 32 } 33 34 35 int main() 36 { 37 // freopen("input.txt","r",stdin); 38 while(cin>>a){ 39 if(a==0) 40 break; 41 else{ 42 s.push(1); 43 bfs(); 44 } 45 } 46 return 0; 47 }