Uva10624-super number (DFS)

Question: uva10624-super number (DFS)

N and m ask you to find such M digits. From the nth bit to the nth bit, the former I bit can be divisible by I. If there is no such number, output-1. if there are multiple, the smallest Lexicographic Order is output.

Solution: enumerate each location with 0 .. 9. Note that the first character cannot be 0, and then DFS determines whether each location meets the requirements. Note that long is exploding here, so we need to modulo it. I thought this would have timed out, and the result would have passed. It is estimated that the number of samples is small, and the number of examples cannot be found.

Code:

#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;}

Uva10624-super number (DFS)