[回溯]Super Number uva10624

Problem B
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 ofi. 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

 

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Problemsetter: Rujia Liu, Member of Elite Problemsetters' Panel

Special Thanks to:

Monirul Hasan (Alternate solution)

Shahriar Manzoor (Figure Drawing)

题意：给定两个数字n，m；求一个长度为m的数字使得这个数字的任意前i(n=<i<=m)个数字组成的数能被i整数。

这道题按照我的思路就是回溯暴力。暂时没想到更好的方法了。

#include<iostream>#include<cstring>using namespace std;int n,m,tag;int arry[50];int mod(int cnt){    int sum=0;    for(int i=0;i<cnt;i++)    {        sum=(sum*10+arry[i])%cnt;    }    return sum;}void dfs(int pos){    if(pos==m)    {        tag=1;        return;    }    if(tag) return;    for(int i=0;i<=9;i++)    {        arry[pos]=i;        if(pos<n-1||(pos>=n-1&&!mod(pos+1)))        {            dfs(pos+1);            if(tag) return;        }    }}int main(){    int k;    cin>>k;    for(int t=1;t<=k;t++)    {        tag=0;        cin>>n>>m;        memset(arry,0,sizeof(arry));        int i,j;        for(i=1;i<=9;i++)        {            arry[0]=i;            dfs(1);            if(tag) break;        }        cout<<"Case "<<t<<": ";        if(tag)        {            for(i=0;i<m;i++) cout<<arry[i];        }        else cout<<"-1";        cout<<endl;    }    return 0;}