fzu 2111 Min Number

 

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

3

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 }

View Code