Let the decimal representation of nn as (x1x2⋯xm)10(x1x2⋯xm)10 satisfying that 1≤x1≤91≤x1≤9, 0≤xi≤90≤xi≤9 (2≤i≤m)(2≤i≤m), which means n=∑mi=1xi10m−in=∑i=1mxi10m−i. In each swap, Anton can select two digits xixi and xjxj (1≤i≤j≤m)(1≤i≤j≤m) and then swap them if the integer after this swap has no leading zero.
Could you please tell him the minimum integer and the maximum integer he can obtain after kk swaps?
InputThe first line contains one integer TT, indicating the number of test cases.
Each of the following TT lines describes a test case and contains two space-separated integers nn and kk.
1≤T≤1001≤T≤100, 1≤n,k≤1091≤n,k≤109.
OutputFor each test case, print in one line the minimum integer and the maximum integer which are separated by one space.
Sample Input
5 12 1 213 2 998244353 1 998244353 2 998244353 3
Sample Output
12 21 123 321 298944353 998544323 238944359 998544332 233944859 998544332
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5
6 using namespace std;
7
8 int maxx,minn,k,len;
9 int c[20],sum1[20],sum2[20],p[20];
10 char ss[20];
11
12 void update()
13 {
14 if(c[ p[1] ]==0)
15 {
16 return;
17 }
18 // 用sum1记录当前排列
19 for(int i=1;i<=len;++i)
20 {
21 sum1[i]=p[i];
22 }
23
24 int kk=0,s=0;
25 for(int i=1;i<=len;++i)
26 {
27 s=s*10 + c[ p[i] ];
28 if(sum1[i] != i)
29 {
30 for(int j=i+1;j<=len;++j)
31 {
32 if(sum1[j]==i)
33 {
34 swap(sum1[i],sum1[j]);
35 ++ kk;
36 // 这一序列不能在k步内实现
37 if(kk>k)
38 {
39 return;
40 }
41 break;
42 }
43 }
44 }
45 }
46
47 // 当前队列满足条件
48 // 更新最大最小值
49 maxx=max(maxx,s);
50 minn=min(minn,s);
51 }
52
53 int main()
54 {
55 int T;
56 scanf("%d",&T);
57 while(T--)
58 {
59 memset(sum1,0,sizeof(sum1));
60 memset(sum2,0,sizeof(sum2));
61 scanf("%s %d",ss,&k);
62
63 len=strlen(ss);
64
65 for(int i=0;i<len;++i)
66 {
67 // 字符转数字
68 c[i+1]=ss[i]-'0';
69 ++ sum1[ c[i+1] ];
70 ++ sum2[ c[i+1] ];
71 }
72
73 // 剪枝
74 if(k>=len-1)
75 {
76 // 输出最小数
77 // 输出第一位（特判非零）
78 for(int i=1;i<=9;++i)
79 {
80 if(sum1[i])
81 {
82 printf("%d",i);
83 -- sum1[i];
84 break;
85 }
86 }
87
88 for(int i=0;i<=9;++i)
89 {
90 while(sum1[i])
91 {
92 printf("%d",i);
93 --sum1[i];
94 }
95 }
96
97 printf(" ");
98
99 // 输出最大数
100 for(int i=9;i>=0;--i)
101 {
102 while(sum2[i])
103 {
104 printf("%d",i);
105 -- sum2[i];
106 }
107 }
108 printf("\n");
109 continue;
110 }
111
112 // 未能剪枝
113
114 // 找一个互不相等的排列（从小到大有序）
115 // 以初始排列为当前数字的编号
116 for(int i=1;i<=len;++i)
117 {
118 p[i]=i;
119 }
120 // 初始化最大最小值
121 minn=2e9;
122 maxx=-1;
123 // 自当前排列开始更新满足条件时的最大最小值
124 do
125 {
126 update();
127 }while(next_permutation(p+1,p+len+1)); // 需要头文件algor
128 // 九位数的全排列，有362880种
129 // 最多尝试次数，为九次
130 // 复杂度约为 10^7
131 printf("%d %d\n",minn,maxx);
132 }
133 return 0;
134 }