Find The Multiple
| Time Limit: 1000MS | Memory Limit: 10000K | |||
| Total Submissions: 30659 | Accepted: 12755 | 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
26190
Sample Output
10100100100100100100111111111111111111
Source
打表。。。
1 #include2 #include 3 #include 4 #include 5 6 using namespace std; 7 8 struct node{ 9 int mod;10 char str[40];11 int cou;12 };13 14 char ans[205][20];15 int i;16 17 void bfs(int n){18 queue que;19 node t;20 t.mod=1;21 t.cou=0;22 t.str[t.cou++]='1';23 24 que.push(t);25 while(!que.empty()){26 node now=que.front();27 que.pop();28 if(now.mod%n==0){29 now.str[now.cou]='\0';30 //strcpy(ans[i],now.str);31 printf("%s,",now.str);32 return ;33 }34 node next=now;35 next.mod=(now.mod*10)%n;36 next.str[next.cou++]='0';37 que.push(next);38 next.mod=(now.mod*10+1)%n;39 next.str[next.cou-1]='1';40 que.push(next);41 }42 }43 44 int main()45 {46 int n;47 for(i=1;i<=200;i++){48 bfs(i);49 }50 51 52 while(scanf("%d",&n)&&n!=0){53 printf("%s\n",ans[n]);54 }55 return 0;56 }
1 #include2 #include 3 #include 4 #include 5 6 using namespace std; 7 8 long long a[220]={ 0,1,10,111,100,10,1110,1001,1000,111111111,10,11,11100,1001,10010,1110,10000,11101,1111111110,11001,100,10101,110,110101,111000,100,10010,1101111111,100100,1101101,1110,111011,100000,111111,111010,10010,11111111100,111,110010,10101,1000,11111,101010,1101101,1100,1111111110,1101010,10011,1110000,1100001,100,100011,100100,100011,11011111110,110,1001000,11001,11011010,11011111,11100,100101,1110110,1111011111,1000000,10010,1111110,1101011,1110100,10000101,10010,10011,111111111000,10001,1110,11100,1100100,1001,101010,10010011,10000,1111111101,111110,101011,1010100,111010,11011010,11010111,11000,11010101,1111111110,1001,11010100,10000011,100110,110010,11100000,11100001,11000010,111111111111111111,100,101,1000110,11100001,1001000,101010,1000110,100010011,110111111100,1001010111,110,111,10010000,1011011,110010,1101010,110110100,10101111111,110111110,100111011,111000,11011,1001010,10001100111,11101100,1000,11110111110,11010011,10000000,100100001,10010,101001,11111100,11101111,11010110,11011111110,11101000,10001,100001010,110110101,100100,10011,100110,1001,1111111110000,11011010,100010,1100001,11100,110111,11100,1110001,11001000,10111110111,10010,1110110,1010100,10101101011,100100110,100011,100000,11101111,11111111010,1010111,1111100,1111110,1010110,11111011,10101000,10111101,111010,1111011111,110110100,1011001101,110101110,100100,110000,100101111,110101010,11010111,11111111100,1001111,10010,100101,110101000,1110,100000110,1001011,1001100,1010111010111,110010,11101111,111000000,11001,111000010,101010,110000100,1101000101,1111111111111111110,111000011,1000}; 9 int main()10 {11 int n;12 while(scanf("%d",&n)&&n!=0){13 printf("%lld\n",a[n]);14 }15 16 return 0;17 }