HDU 3183.A Magic Lamp【RMQ】【4月18】

A Magic Lamp

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3022    Accepted Submission(s): 1174

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

原來這就是RMQ~給一個長度不超過1000位的數，使得刪去m個數字之後剩下的部分組成的數字最小。

思路：

長度爲l的數，刪去m個數字之後還剩l-m位，即從l位數中選取l-m位數字使得組成的數字最小。

按順序從最高位取，第1~m+1位中取最高位，記錄取得位置爲pos，則下一次從pos+1~m+2中取下一位，以此類推：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
const int MAX = 1010;
char n[MAX];
int m, l, pos, maxn, stpos, first;
int main()
{
    while(scanf("%s %d", n, &m) != EOF)
    {
        first = 1;
        l = strlen(n);
        int len = l - m;
        stpos = 0;
        if(l == m)
        {
            cout << 0 << endl;
            continue;
        }
        for(int i = 0;i < len; ++i)
        {
            maxn = 10;
            for(int j = stpos;j <= m+i; ++j)
            {
                if(n[j] - '0' < maxn)
                {
                    maxn = n[j]-'0';
                    pos = j;//記錄位置
                }
            }
            if(first == 0 || i == len-1) cout << n[pos];
            else if(n[pos] != '0')//前導0不要輸出
            {
                first = 0;
                cout << n[pos];
            }
            stpos = pos+1;
        }
        cout << endl;
    }
    return 0;
}