E - Perfect Number

标签：its   tmp   超过   tin   find   bsp   tle   not   one   

Problem description

We consider a positive integer perfect, if and only if the sum of its digits is exactly 10. Given a positive integer k, your task is to find the k-th smallest perfect positive integer.

Input

A single line with a positive integer k (1?≤?k?≤?10?000).

Output

A single number, denoting the k-th smallest perfect integer.

Examples

1

19

2

28

Note

The first perfect integer is 19 and the second one is 28.

解题思路：题目的意思就是找出升序序列（每个数的每位数字总和都为10）中第k个数。暴力水过！

AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int n,k=0,tmp,sum,obj[10005];bool flag;
 6     for(int i=1;k<10000;++i){
 7         tmp=i;sum=0;flag=false;
 8         while(!flag && tmp){
 9             if(sum>=10)flag=true;//如果sum超过10就无需再比较下去了
10             sum+=tmp%10;
11             tmp/=10;
12         }
13         if(!flag && sum==10)obj[k++]=i;
14     }
15     cin>>n;
16     cout<<obj[n-1]<<endl;
17     return 0;
18 }

E - Perfect Number

标签：its   tmp   超过   tin   find   bsp   tle   not   one   

原文地址：https://www.cnblogs.com/acgoto/p/9104977.html