dfs - 卡一个无符号长整形

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 ： 对于给定的一个数 ， 输出的一个数是他的倍数，并且十进制位只有 0 和 1 

思路 ： DFS 和 BFS 都可以做 
　　bfs 的代码

#define ll unsigned long long

int n;
void bfs(){
    queue<ll>que;
    que.push(1);

    while(!que.empty()){
        ll a = que.front();
        que.pop();

        if (a % n == 0) {
            printf("%lld\n", a);
            break;
        }
        que.push(a*10);
        que.push(a*10+1);
    }
}

int main() {

    while(~scanf("%d", &n) && n){
        bfs();
    }
    //for(int i = 1; i <= 200; i++){
        //n = i;
        //bfs();
    //}
    return 0;
}

　　而用dfs 做 ， 考虑好递归结束的条件，如果这题用 long long 的话就直接 wa 了，有些时候要想想 unsigned 的强大地方可以对长整型变量的精度增加一位。

dfs 的代码 ：

　　

#define ll unsigned long long
ll n;
int sign;

void dfs(ll num, int cnt){
    if (cnt > 19 || sign) return;
    if (num % n == 0) {
        printf("%lld\n", num);
        sign = 1;
        return;  // 递归函数里要写 return ， 不能写 break
    }
    dfs(num*10, cnt+1);
    dfs(num*10+1, cnt+1);
}

int main() {

    while(~scanf("%lld", &n) && n){
        sign = 0;
        dfs(1, 1);
    }
    return 0;
}