hdu 3183 A Magic Lamp

A Magic Lamp

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2954    Accepted Submission(s): 1152

Problem Description

Kiki likes traveling. One day she finds a magic lamp, unfortunately the genie in the lamp is not so kind. Kiki must answer a question, and then the genie will realize one of her dreams.
The question is: give you an integer, you are allowed to delete exactly m digits. The left digits will form a new integer. You should make it minimum.
You are not allowed to change the order of the digits. Now can you help Kiki to realize her dream?

 

Input

There are several test cases.
Each test case will contain an integer you are given (which may at most contains 1000 digits.) and the integer m (if the integer contains n digits, m will not bigger then n). The given integer will not contain leading zero.

 

Output

For each case, output the minimum result you can get in one line.
If the result contains leading zero, ignore it.

 

Sample Input

  

   178543 4 
1000001 1
100001 2
12345 2
54321 2
  

 

Sample Output

  

   13
1
0
123
321
  

 

Source

HDU 2009-11 Programming Contest

 

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3183

题目大意：给个数字，求去掉m位数字后得到的数字最大。

解题思路：思维题，数字长度范围1000，用数组存储，因为要去掉m个数字，相当于选出n-m个数字，可以发现，选第一位时，为了保证后面至少有n-m-1个数字能继续选择，第一位选择的数字下标范围在时0~m，从中选出最小的数字，记录下标st,选第二位数字时下标范围为(st+1)~(m+1)，依次类推。

代码如下：

#include <cstdio>
#include <cstring>
const int maxn=1000+10;
int a[maxn],ans[maxn];
char s[maxn];
int main(void)
{
	int n,m;
	while(scanf("%s",s)!=EOF)
	{
		memset(a,0,sizeof(a));
		memset(ans,0,sizeof(ans));
		scanf("%d",&m);
		n=strlen(s);
		for(int i=0;i<n;i++)
			a[i]=s[i]-'0';
		int st=0,ed=m;
		for(int i=0;i<n-m;i++)
		{
			ans[i]=9;
			for(int j=st;j<=ed;j++)
			{
				if(ans[i]>a[j])
				{
					ans[i]=a[j];
					st=j+1;
				}
			}
			ed++;
		}
		int j=n-m-1;
		for(int i=0;i<n-m;i++)
			if(ans[i])
			{
				j=i;
				break;
			}
		for(int i=j;i<n-m;i++)
			printf("%d", ans[i]);
		printf("\n");
	}
}