Find The Multiple
Time Limit: 1000MS | Memory Limit: 10000K | |||
Total Submissions: 28780 | Accepted: 11935 | 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
2 6 19 0
Sample Output
10 100100100100100100 111111111111111111大意是给出一个数n,求出n的一个仅含0或1的倍数。
这是稍微暴力地深搜
# include <stdio.h> int n, flag; unsigned long long result; void dfs(int step, unsigned long long sum) { int i; if(sum % n == 0) { flag = 1; result = sum; return; } if(step == 19) return; for(i=0; i<=1; ++i) { sum = sum*10 + i; dfs(step+1, sum); sum = (sum-i)/10; if(flag) return; } } int main() { while(scanf("%d",&n), n) { flag = 0; dfs(1, 1); printf("%llu\n",result); } return 0; }
这是广搜的做法
# include <stdio.h> # include <string.h> struct node { int num, pre, sum;//pre记录父节点的下标 }a[300]; int main() { int b[300], vis[300], n, i, k, temp, head, tail, flag, result; while(scanf("%d",&n),n) { memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b)); memset(vis, 0, sizeof(vis)); k = flag = head = tail = 0; a[head].num = 1; a[head].sum = 1; a[head].pre = -1; while(head <= tail) { for(i=0; i<=1; ++i) { temp = (a[head].sum * 10 + i)%n;//同余定理 if(!vis[temp])//★排除重复的情况 { ++tail; a[tail].pre = head; a[tail].num = i; a[tail].sum = temp; vis[temp] = 1; } if(a[tail].sum == 0) { result = tail; flag = 1; break; } } if(flag) break; ++head; } while(a[result].pre != -1)//将路线节点的下标找出来 { b[k++] = a[result].pre; result = a[result].pre; } --k; for(i=k; i>=0; --i)//将找出来的下标输出对应的值 { printf("%d",a[b[i]].num); } printf("%d\n",a[tail].num); } return 0; }