codeforces--round#634 B-Construct the String

Construct the String

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.

You have to answer t independent test cases.

Recall that the substring s[l…r] is the string sl,sl+1,…,sr and its length is r−l+1. In this problem you are only interested in substrings of length a.

Input
The first line of the input contains one integer t (1≤t≤2000) — the number of test cases. Then t test cases follow.

The only line of a test case contains three space-separated integers n, a and b (1≤a≤n≤2000,1≤b≤min(26,a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.

It is guaranteed that the sum of n over all test cases does not exceed 2000 (∑n≤2000).

Output
For each test case, print the answer — such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.

**Example
input
4
7 5 3
6 1 1
6 6 1
5 2 2
output
tleelte
qwerty
vvvvvv
abcde

Note
In the first test case of the example, consider all the substrings of length 5:

“tleel”: it contains 3 distinct (unique) letters,
“leelt”: it contains 3 distinct (unique) letters,
“eelte”: it contains 3 distinct (unique) letters.

AC的C++代码如下：

#include<iostream>
#include<string>
using namespace std;
char str[26] = { 'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z' };

int main()
{
	int t, n, a, b;
	string s;
	cin >> t;
	while (t--)
	{
		cin >> n >> a >> b;
		int i, r;
		s.clear();
		for (r = 0; r < b; r++)
			s.push_back(str[r]);
		for (i = 0; i < a - b; i++)
		{
			s.push_back(str[r - 1]);
		}
		string l = s;
		for (int i = 1; i < n / a; i++)
			s += l;
		if (s.size() != n)
		{
			string l = s.substr(0, n - s.size());
			s += l;
		}

		cout << s << endl;
	}


}