Anton has a positive integer nn, however, it quite looks like a mess, so he wants to make it beautiful after kk swaps of digits.
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?
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?
Input
The first line contains one integer T indicating the number of test cases.
Each of the following T lines describes a test case and contains two space-separated integers n and k.
1≤T≤1001≤T≤100, 1≤n,k≤1091≤n,k≤109.
Output
For 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
题意
给一个数 N 每次可以交换两个数字的位置
文交换 K 次后 可以得到的最大值和最小值
解法
应该老老实实开数组作各位分离写数据结构，
但是脑子抽了筋偏要用算术运算的方法来实现交换位。
我下次再也不瞎鸡乱写了。
大概思路
从高位开始向低位扫描，
比较当前位与最大/最小值，
如果有必要交换的话，
每次把数中合法的（首位不为零）最大/最小数提到当前位置。
错误代码（至今不知道wa在哪里，也不想改了）
1 #include <bits/stdc++.h> 2 #include <iostream> 3 #include <cstring> 4 #include <stack> 5 #include <cstdlib> 6 #include <queue> 7 #include <cmath> 8 #include <cstdio> 9 #include <algorithm> 10 #include <string> 11 #include <vector> 12 #include <list> 13 #include <iterator> 14 #include <set> 15 #include <map> 16 #include <utility> 17 #include <iomanip> 18 #include <ctime> 19 #include <sstream> 20 #include <bitset> 21 #include <deque> 22 #include <limits> 23 #include <numeric> 24 #include <functional> 25 #include <ctime> 26 27 #define gc getchar() 28 #define mem(a) memset(a,0,sizeof(a)) 29 #define mod 1000000007 30 #define sort(a,n,int) sort(a,a+n,less<int>()) 31 #define fread() freopen("in.in","r",stdin) 32 #define fwrite() freopen("out.out","w",stdout) 33 using namespace std; 34 35 typedef long long ll; 36 typedef char ch; 37 typedef double db; 38 39 int gcd(int a,int b){ 40 if(a<b)swap(a,b); 41 if(a%b==0)return b; 42 else gcd(b,a%b); 43 } 44 45 double random(double start,double end) 46 { 47 return start + (end - start) * rand() / (RAND_MAX + 1.0); 48 } 49 const int maxn = 100+10; 50 const int INF = 0x3f3f3f3f; 51 double dis(double x1,double y1){ 52 return sqrt(x1 * x1 + y1 * y1); 53 } 54 55 int main() 56 { 57 ll t = 0; 58 cin >> t; 59 while(t--) 60 { 61 unsigned long long a = 0 , k = 0; 62 unsigned long long res = 0; 63 cin >> a >> k; 64 //cout<<t<<endl;a = t;k = 2; 65 res = a; 66 unsigned long long a1 = a; 67 int j = 0; 68 for(int i = 0 , counter = 0;counter<k;) 69 { 70 unsigned long long s = a; 71 int min = 9; 72 int p = -1; 73 for(j = 0;s/10 != 0;j++)s/=10; 74 //cout<<j<<endl; // 75 if(j==i)break; 76 if(j==1) 77 { 78 if(a%10<a/10 && a%10!=0) 79 res = (a%10) * 10 + a/10; 80 else res = a; 81 break; 82 } 83 s = a; 84 for(int l = 0;l<j-i;l++) 85 { 86 if(s % 10 < min) 87 { 88 if(i!=0 || s % 10!=0) 89 { 90 min = s % 10; 91 p = l; 92 } 93 } 94 s /= 10; 95 } 96 if(p==-1) 97 { 98 res = a; 99 break; 100 } 101 //cout<<i<<' '<<counter<<endl;// 102 //cout<<p<<endl; // 103 s = a; 104 int m = 0,n = 0; 105 int k = 0; 106 107 for(k = 0;k <=j-i;k++) 108 { 109 if(k == p) 110 { 111 m = s % 10; 112 } 113 if(k == j-i) 114 { 115 n = s % 10; 116 } 117 s/=10; 118 } 119 //cout<<m<<' '<<n<<endl; // 120 if(n <= m) 121 { 122 i++; 123 continue; 124 } 125 int mul = 1; 126 for(k = 0;k <=j-i;k++) 127 { 128 if(k == p) 129 { 130 a = a- m*mul +n*mul; 131 } 132 if(k == j-i) 133 { 134 a = a- n*mul +m*mul; 135 } 136 mul*=10; 137 } 138 res = a; 139 //cout<<a<<endl; // 140 i++; 141 counter++; 142 } 143 cout<<res<<' '; 144 a = a1; 145 res = a; 146 for(int i = 0 , counter = 0;counter<k;) 147 { 148 if(i>j) 149 { 150 res = a; 151 break; 152 } 153 ll s = a; 154 int max = 0; 155 int p = 0; 156 for(j = 0;s/10 != 0;j++)s/=10; 157 //cout<<j<<endl; // 158 if(j==i)break; 159 if(j==1) 160 { 161 if(a%10>a/10 && a%10!=0) 162 res = (a%10)*10+a/10; 163 else res = a; 164 break; 165 } 166 167 s = a; 168 for(int l = 0;l<j-i;l++) 169 { 170 if(s % 10 > max) 171 { 172 max = s % 10; 173 p = l; 174 } 175 s /= 10; 176 } 177 //cout<<p<<endl; // 178 s = a; 179 int m = 0,n = 0; 180 int k = 0; 181 //cout<<s<<endl; // 182 for(k = 0;k <=j-i;k++) 183 { 184 if(k == p) 185 { 186 m = s % 10; 187 } 188 if(k == j-i) 189 { 190 n = s % 10; 191 } 192 s/=10; 193 } 194 //cout<<m<<' '<<n<<endl; // 195 if(n >= m) 196 { 197 i++; 198 continue; 199 } 200 int mul = 1; 201 for(k = 0;k <=j-i;k++) 202 { 203 if(k == p) 204 { 205 a = a- m*mul +n*mul; 206 } 207 if(k == j-i) 208 { 209 a = a- n*mul +m*mul; 210 } 211 mul*=10; 212 } 213 res = a; 214 i++; 215 counter++; 216 } 217 cout<<res<<endl; 218 219 } 220 return 0; 221 }