[编程题]小易的字典

链接：https://www.nowcoder.com/questionTerminal/12b1b8ef17e1441f86f322b250bff4c0?source=relative
来源：牛客网

小易在学校中学习了关于字符串的理论, 于是他基于此完成了一个字典的项目。

小易的这个字典很奇特, 字典内的每个单词都包含n个’a’和m个’z’, 并且所有单词按照字典序排列。

小易现在希望你能帮他找出第k个单词是什么。

输入描述:
输入包括一行三个整数n, m, k(1 <= n, m <= 100, 1 <= k <= 109), 以空格分割。

输出描述:
输出第k个字典中的字符串，如果无解，输出-1。
示例1
输入
2 2 6
输出
zzaa
说明
字典中的字符串依次为aazz azaz azza zaaz zaza zzaa

fun函数数为了求排列组合的。

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        long k = sc.nextInt();

        char[] res = new char[m + n];
        for(int i = 0; i < m + n; ++i){
            res[i] = 'z';
        }

        int xuan = n - 1;
        int total = m + n - 1;
        long sum_temp = 0;
        boolean flag_while = true;
        while(true){
            long temp = fun(xuan, total);
            //long temp = pz(xuan, total - xuan, k);
            sum_temp += temp;
            if(sum_temp < k){
                --total;
                if(total < xuan){
                    flag_while = false;
                    break;
                }
            }else if(sum_temp == k){
                res[m + n - total - 1] = 'a';
                for(int i = 0; i < xuan; ++i){
                    res[m + n - 1 - i] = 'a';
                }
                break;
            }else{
                sum_temp -= temp;
                res[m + n - total - 1] = 'a';
                --xuan;
                --total;
                if(total < xuan){
                    flag_while = false;
                    break;
                }
            }
        }

        if(flag_while == true){
            for(int i = 0; i < res.length; ++i){
                System.out.printf("%c", res[i]);
            }
        }else{
            System.out.printf("-1");
        }
    }

    public static long fun(int xuan, int total){
        double fenmu = 1;
        int size = Math.min(xuan, total - xuan);

        for(int i = 0; i < size; ++i){
            fenmu *= (total - i);
            fenmu /= (i + 1);
        }
        return (long)fenmu;
    }
}

但是如果讲fun函数稍稍改动一下，结果就不对了

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        long k = sc.nextInt();

        char[] res = new char[m + n];
        for(int i = 0; i < m + n; ++i){
            res[i] = 'z';
        }

        int xuan = n - 1;
        int total = m + n - 1;
        long sum_temp = 0;
        boolean flag_while = true;
        while(true){
            long temp = fun(xuan, total);
            //long temp = pz(xuan, total - xuan, k);
            sum_temp += temp;
            if(sum_temp < k){
                --total;
                if(total < xuan){
                    flag_while = false;
                    break;
                }
            }else if(sum_temp == k){
                res[m + n - total - 1] = 'a';
                for(int i = 0; i < xuan; ++i){
                    res[m + n - 1 - i] = 'a';
                }
                break;
            }else{
                sum_temp -= temp;
                res[m + n - total - 1] = 'a';
                --xuan;
                --total;
                if(total < xuan){
                    flag_while = false;
                    break;
                }
            }
        }

        if(flag_while == true){
            for(int i = 0; i < res.length; ++i){
                System.out.printf("%c", res[i]);
            }
        }else{
            System.out.printf("-1");
        }
    }

    public static long fun(int xuan, int total){
        double fenmu = 1;
        int size = Math.min(xuan, total - xuan);

        for(int i = 0; i < size; ++i){
            fenmu *= (total - i) / (i + 1);
        }
        return (long)fenmu;
    }
}

不太清楚这是为什么