UVA - 10624 Super Number（回溯）

Super Number

Input: Standard Input
Output: Standard Output
Time Limit: 3 Seconds


Don't you think 162456723 very special? Look at the picture below if you are unable to find its speciality. (a | b means �b is divisible by a�)



Figure: Super Numbers


Given n, m (0 < n < m < 30), you are to find a m-digit positive integer X such that for every i (n <= i <= m), the first i digits of X is a multiple of i.
If more than one such X exists, you should output the lexicographically smallest one. Note that the first digit of X should not be 0.

Input

The first line of the input contains a single integer t(1 <= t <= 15), the number of test cases followed. For each case, two integers n and m are
separated by a single space.

Output

For each test case, print the case number and X. If no such number, print -1.

Sample Input�������������������������� Output for Sample Input

2 1 10 3 29

Case 1: 1020005640 Case 2: -1

题目大意：

要求你求出一个超级数字，超级数字的定义如下：

由m位，所有前i位，都能被数字i整除（n <= i <= m），如果超级数字不止一个要求输出字典序最小的超级数字。

注意：超级数字的第一位不能为0

举个例子：

输入

1 10

输出

1020005640

1 % 1 = 0

10 % 2 = 0

102 % 3 = 0

1020 % 4 = 0

10200 % 5 = 0

102000 % 6 = 0

1020005 % 7 = 0

...

1020005640 % 10 = 0

所以当n=1，m=10时，1020005640是超级数字。

解析：数字最多有30位需要用数组进行保存，回溯求解，先1~9，枚举出第一位，再用回溯枚举出剩下的位数，每个位数的值为0~9，注意当cur < n时可以直接枚举下一位，当cur >= n时，需要判断前cur位是否能被cur整除，但是这个数字可能非常大（最多30位），所以要模拟除法的过程，步骤如下：

1.从首位开始对cur求余数

2.(余数*10+下一位) % cur

3.重复2的操作，直到当前的位数 == cur结束。

#include <cstdio>
#include <cstring>
using namespace std;
bool ok;
int m,n;
int num[35];
inline int judge(int cur) {
	int sum = 0;
	for(int i = 1; i <= cur; i++) {
		sum = (sum * 10 + num[i]) % cur;
	}
	return sum;
}
inline void dfs(int cur) {
	if(cur == m+1) {
		ok = true;
		return ;
	}
	for(int i = 0; i < 10; i++) {
		num[cur] = i;
		if(cur < n || (cur >= n && !judge(cur)) ) {
			dfs(cur+1);
		}
		if(ok) {
			return ;
		}
	}
}
int main() {
	int t;
	int cas = 1;
	scanf("%d",&t);
	while(t--) {
		scanf("%d%d",&n,&m);
		for(int i = 1; i < 10; i++) {
			num[1] = i;
			ok = false;
			dfs(2);
			if(ok) {
				break;
			}
		}
		printf("Case %d: ",cas++);
		if(ok) {
			for(int i = 1; i <= m; i++) {
				printf("%d",num[i]);
			}
			printf("\n");
		}else {
			printf("-1\n");
		}
	}
	return 0;
}