Educational Codeforces Round 21 A

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本文介绍了一个简单的算法，用于计算从当前年份到下一个幸运年的年数。幸运年是指一个年份的十进制表示中非0数字不超过一个的年份。文章通过示例解释了如何确定下一个幸运年，并提供了一段C++代码实现。

A. Lucky Year

Apart from having lots of holidays throughout the year, residents of Berland also have whole lucky years. Year is considered lucky if it has no more than 1 non-zero digit in its number. So years 100, 40000, 5 are lucky and 12, 3001 and 12345 are not.

You are given current year in Berland. Your task is to find how long will residents of Berland wait till the next lucky year.

Input
The first line contains integer number n (1 ≤ n ≤ 10^9) — current year in Berland.

Output
Output amount of years from the current year to the next lucky one.

Examples

input
4
output
1

input
201
output
99

input
4000
output
1000

Note
In the first example next lucky year is 5. In the second one — 300. In the third — 5000.

大致题意：当一个十进制表示的年份n中各个位上的非0数不超过1个时，我们把该年份称为幸运年，告诉你现在的年份，问最少还需过多少年才是幸运年

思路：水题，直接看代码吧。

代码如下

#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<cstring>
#include<cmath>
#define LL long long  
#define ULL unsigned long long  
using namespace std;

int main()
{
      int n,n1;
      cin>>n;
      if(n<10)
      printf("1");
      else 
      {
        int i=1;
        while(i<=n)
        {
            i*=10;
        }
         i/=10;
         if(n%i==0)
         {
            printf("%d",i);
          } 
          else 
          printf("%d",(n/i+1)*i-n);
      }
    return 0; 
}