Codeforces Round #634 B. Construct the String（贪心）

题目描述

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

You have to answer t t t independent test cases.

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

输入格式

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

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

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

输出格式

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

输入输出样例

输入 #1 复制

4
7 5 3
6 1 1
6 6 1
5 2 2

输出 #1 复制

tleelte
qwerty
vvvvvv
abcde

说明/提示

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

• "tleel": it contains 3 3 3 distinct (unique) letters,
• "leelt": it contains 3 3 3 distinct (unique) letters,
• "eelte": it contains 3 3 3 distinct (unique) letters.

直接构造循环节长度为b的字符串，比如n=7 b=3 就构造abcabca 这样能在尽可能短的距离内让不同字符出现的次数比较均匀。和a没啥关系。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,a,b;
        scanf("%d%d%d",&n,&a,&b);
        int i,j;
        for(i=0;i<n;i++)
        {
            printf("%c",i%b+'a');
        }
        cout<<endl;
    }
}