fzu 2111 Min Number

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原文链接：http://www.cnblogs.com/jeff-wgc/p/4449377.html

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本文介绍了一种通过字符串操作来获取给定10进制数在特定操作次数下的最小可能值的方法。通过选择合适的字符进行交换，实现从原始数到最小数的转换。

http://acm.fzu.edu.cn/problem.php?pid=2111

Problem 2111 Min Number

Accept: 572 Submit: 1106
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

Now you are given one non-negative integer n in 10-base notation, it will only contain digits ('0'-'9'). You are allowed to choose 2 integers i and j, such that: i!=j, 1≤i<j≤|n|, here |n| means the length of n’s 10-base notation. Then we can swap n[i] and n[j].

For example, n=9012, we choose i=1, j=3, then we swap n[1] and n[3], then we get 1092, which is smaller than the original n.

Now you are allowed to operate at most M times, so what is the smallest number you can get after the operation(s)?

Please note that in this problem, leading zero is not allowed!

Input

The first line of the input contains an integer T (T≤100), indicating the number of test cases.

Then T cases, for any case, only 2 integers n and M (0≤n<10^1000, 0≤M≤100) in a single line.

Output

For each test case, output the minimum number we can get after no more than M operations.

Sample Input

9012 0

9012 1

9012 2

Sample Output

9012

1092

1029

分析：

由于数字较大10^100 , 所以考虑字符串解决，只需判断是否为首字符，是的话和后面的最小的靠后的非‘0’字符交换，否的话和后面的最小的字符交换即可。

AC代码：

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <stdlib.h>
 4 #include <algorithm>
 5 #include <string.h>
 6 #include <string>
 7 #include <math.h>
 8 #include <map>
 9 #include <set>
10 #include <vector>
11 #include <stack>
12 #include <queue>
13 
14 using namespace std;
15 
16 const int INF = 0x3f3f3f3f;
17 const int MAX = 100 + 10;
18 const double eps = 1e-7;
19 const double PI = acos(-1.0);
20 
21 char str[MAX];
22 int len;
23 
24 int judge(int n)
25 {
26     int temp = n , i;
27     if(n == 0)
28     {
29         char min = str[n];
30         for(i = 1;i < len;i ++)
31         {
32             if(str[i] != '0' && str[i] <= min)
33             {
34                 min = str[i];
35                 temp = i;
36             }
37         }
38     }
39     else
40     {
41         char min = str[n];
42         for(i = n + 1;i < len ;i ++)
43         {
44             if(str[i] <= min)
45             {
46                 min = str[i];
47                 temp = i;
48             }
49         }
50     }
51     return temp;
52 }
53 
54 int main()
55 {
56     int T , n;
57     scanf("%d",&T);
58     while(T --)
59     {
60         scanf("%s %d",str , &n);
61         len = strlen(str);
62         int i = 0;
63         while(n --)
64         {
65             int ji = judge(i);
66             if(ji == i)
67             {
68                 n ++;
69                 i ++;
70             }
71             else
72             {
73                 str[i] = (str[ji] ^ str[i] ^ (str[ji] = str[i]));
74                 i ++;
75             }
76             if(i == len)
77                 break;
78         }
79         for(i = 0;i < len ;i ++)
80             printf("%c",str[i]);
81         puts("");
82     }
83     return 0;
84 }