UVA10624 - Super Number(dfs)

题目：UVA10624 - Super Number(dfs)

题目大意：给你n和m要求找出这样的m位数，从第n位到第m位都满足前i位是可以被i整除，如果没有这样的数，输出-1.有多个就输出字典序最小的那个。

解题思路：将每个位置都用0..9枚举一下，注意第一个字符不能是0，然后dfs判断每个位置是否都满足要求。注意这里是会爆long long的，所以要取模一下。本来以为这样的做法会超时的，结果竟然过了，估计是样例数少，而且找不到的情况也比较少。

代码：

#include <cstdio>#include <cstring>const int N = 35;typedef unsigned long long ll;int num[N];int m, n;bool judge (int cur) {ll sum = 0;for (int i = 0; i <= cur; i++) {sum = sum * 10 + num[i];if (i > 17)sum = sum % (cur + 1);}sum = sum % (cur + 1);return sum ? false : true;}bool dfs (int cur) {if (cur == m) return true;for (int i = 0; i < 10; i++) {if (!cur && !i)continue;num[cur] = i;if (cur < n - 1 || judge(cur)) {if (dfs (cur + 1))return true;}}return false;}int main () {int t;scanf ("%d", &t);for (int i = 1; i <= t; i++) {scanf ("%d%d", &n, &m);printf ("Case %d: ", i);if (!dfs(0)) printf ("-1\n");else {for (int j = 0; j < m; j++)printf ("%d", num[j]);printf ("\n");}}return 0;}