https://codeforces.com/contest/1811/problem/A
以上是本题地址。
You have a positive number of length n and one additional digit.
You can insert this digit anywhere in the number,
including at the beginning or at the end.
Your task is to make the result as large as possible.
For example,
you have the number 76543,and the additional digit is 4.
Then the maximum number you can get is 765443,and it can be obtained in two ways:
by inserting a digit after the 3th or after the 4th digit of the number.
Input
The first line contains a single integer t(1 <= t <= 104) - the number of test cases.
The descriptions of the test cases follow.
The first line of the description of each test case contains two
integers N and D(1 <= n <= 2 * 100000; 0 <= d <= 9) - the length of the number
and an additional digit ,respectively.
The second line of the description of each test case contains a string consisting of n digits-
the number that you have inintially.
It is guaranteed that the number does not contain leading zeros.
It is guaranteed that the number does not contain leading zeros.
It is guaranteed that the sum of n for all test cases does not exceed 2*100000
*/
/*
Example
Input
11
5 4
76543
1 0
1
2 5
44
3 6
666
5 6
13579
5 8
97531
19 4
9876543210123456789
5 7
73737
8 1
20000000
7 0
7058959
12 1
828127127732
Output
765443
10
544
6666
613579
987531
98765443210123456789
773737
210000000
70589590
8281271277321
Note:
Also you can input one case and output one result.
根据题意：
将一个数字插入到一个字符串里，需要满足的条件是：使得字符串数字变得最大。
注意：字符串数字的里面的顺序不可以进行移动，只能插入最新的数据。
思路：将数据先插入到头，然后比较和后面的数字，如果大，就不比较了，如果小，就移动这个数据到下一个，循环，直到这个数字>后面的数字，就停止。
在codeforces上测试，AC100
以下是代码：
cpp
/*
You have a positive number of length n and one additional digit.
You can insert this digit anywhere in the number,
including at the beginning or at the end.
Your task is to make the result as large as possible.
For example,
you have the number 76543,and the additional digit is 4.
Then the maximum number you can get is 765443,and it can be obtained in two ways:
by inserting a digit after the 3th or after the 4th digit of the number.
Input
The first line contains a single integer t(1 <= t <= 104) - the number of test cases.
The descriptions of the test cases follow.
The first line of the description of each test case contains two
integers N and D(1 <= n <= 2 * 100000; 0 <= d <= 9) - the length of the number
and an additional digit ,respectively.
The second line of the description of each test case contains a string consisting of n digits-
the number that you have inintially.
It is guaranteed that the number does not contain leading zeros.
It is guaranteed that the number does not contain leading zeros.
It is guaranteed that the sum of n for all test cases does not exceed 2*100000
*/
/*
Example
Input
11
5 4
76543
1 0
1
2 5
44
3 6
666
5 6
13579
5 8
97531
19 4
9876543210123456789
5 7
73737
8 1
20000000
7 0
7058959
12 1
828127127732
Output
765443
10
544
6666
613579
987531
98765443210123456789
773737
210000000
70589590
8281271277321
Note:
Also you can input one case and output one result.
*/
#include <iostream>
#include <string>
#include <vector>
using namespace std;
void move_to_good_pos(string &str1, int d)
{
string a = to_string(d);
str1 = a+str1;
int len = str1.length();
for (int i = 0; i < len-1; i++)
{
if (a[0]>str1[i + 1])
{
break;
}
if (str1[i] < str1[i + 1])
{
swap(str1[i], str1[i + 1]);
}
}
}
string insert_digit()
{
string str1;
int n, d;
cin >> n >> d;
getchar();
getline(cin, str1);
move_to_good_pos(str1,d);
return str1;
}
int main()
{
int t = 0;
cin >> t;
int pos = 0;
vector<string> list;
string str1="";
while (t--)
{
pos++;
str1=insert_digit();
list.push_back(str1);
}
for (int i = 0; i < list.size(); i++)
{
cout << list[i] << endl;
}
return 0;
}