9月13日微众银行笔试代码
1.拼接数字：数字统一转成String,后续处理比较方便，首先按字符串长度和字典序排序，从里面找出最大的三个数存到另一个数组中，然后按字典序排序，最后拼接即可得到答案
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
String[] nums = new String[n];
for (int i = 0; i < n; i++) {
nums[i] = sc.next();
}
Arrays.sort(nums, (o1, o2) -> {
if ( o1.length()==o2.length() ) {
return o1.compareTo(o2);
} else
return o1.length()-o2.length();
});
String[] temp = new String[]{nums[n-1],nums[n-2],nums[n-3]};
Arrays.sort(temp);
StringBuilder ans = new StringBuilder();
for (int i = temp.length - 1; i >= 0; i--) {
ans.append(temp[i]);
}
System.out.println(ans);
}
} 2.最少变换次数：如果能够变换，则较大的数一定是较小的数的2次幂，找到这个倍数关系即可轻松解答 import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
long a,b;
while ( n-->0 ) {
int ans = Integer.MAX_VALUE;
a = sc.nextLong();
b = sc.nextLong();
long x = Math.max(a,b);
long y = Math.min(a,b);
if ( x==y )
ans = 0;
// 判断能否整除
else if ( x%y!=0 )
ans = -1;
else {
// 判断x是否是y的2次方
long t = x/y;
int cnt = 0;
while ( t>1 ) {
if ( t%2==1 ) {
ans = -1;
break;
}
t /= 2;
cnt++;
}
if ( ans!=-1 ) {
ans = 0;
ans += cnt/3;
cnt %= 3;
ans += cnt/2;
cnt %= 2;
ans += cnt;
}
}
System.out.println(ans);
}
}
} 3.上升子序列：（未AC，纯暴力+剪枝） import java.util.*;
public class Main {
static long ans = 0,mod = 998244353;
static int[] dp;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n,m;
n = sc.nextInt();
m = sc.nextInt();
dp = new int[n];
dfs(new int[n],0,n,m);
System.out.println(ans%mod);
}
public static void dfs( int[] nums,int index,int n,int m ) {
int length = lis(nums,index);
if ( length>3 )
return;
if ( n-index<3-length )
return;
if ( index==n ) {
if ( length==3 )
ans++;
return;
}
for (int i = 1; i <= m; i++) {
nums[index] = i;
dfs(nums,index+1,n,m);
}
}
// 求最长上升子序列
public static int lis( int[] nums,int cnt ) {
if ( cnt<2 )
return cnt;
int res = 1;
Arrays.fill(dp,1);
for (int i = 1; i < cnt; i++) {
for (int j = 0; j < i; j++) {
if ( nums[i]>nums[j] ) {
dp[i] = Math.max(dp[j]+1,dp[i]);
}
}
res = Math.max(res,dp[i]);
}
return res;
}
} 
