fzu 2111 Min Number-白红宇

fzu 2111 Min Number

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发布时间：2019-06-12

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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 
         
           2 #include 
          
            3 #include 
           
             4 #include 
            
              5 #include 
             
               6 #include 
              
                7 #include 
               
                 8 #include 
                 9 #include 
                 
                  10 #include 
                  
                   11 #include 
                   
                    12 #include 
                    
                     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 else40 {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 else72 {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 }

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转载于:https://www.cnblogs.com/jeff-wgc/p/4449377.html