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’)

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
For each test case, print the case number and X. If no such number, print -1.
|
2 1 10 3 29 |
Case 1: 1020005640 Case 2: -1
|
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;
}
原文:http://blog.csdn.net/shaerdong/article/details/19417241