codeforces 835B The number on the board （优先队列，贪心）

最新推荐文章于 2023-01-21 17:35:43 发布

原创最新推荐文章于 2023-01-21 17:35:43 发布 · 467 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#贪心

水题专栏收录该内容

11 篇文章

订阅专栏

本文介绍了一个算法问题：给定一个数n和一个数字之和k，原板上有一个自然数其数字之和不小于k，经过修改变为n。在长度不变的情况下，文章探讨了如何找到原始数字与n之间的最小不同位数，并提供了具体的实现代码。

Some natural number was written on the board. Its sum of digits was not less than k. But you were distracted a bit, and someone changed this number to n, replacing some digits with others. It's known that the length of the number didn't change.

You have to find the minimum number of digits in which these two numbers can differ.

Input

The first line contains integer k (1 ≤ k ≤ 109).

The second line contains integer n (1 ≤ n < 10100000).

There are no leading zeros in n. It's guaranteed that this situation is possible.

Output

Print the minimum number of digits in which the initial number and n can differ.

Examples

input
3
11

output
1

input
3
99

output
0

Note

In the first example, the initial number could be 12.

In the second example the sum of the digits of n is not less than k. The initial number could be equal to n.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1000100;

int k,sum,ans;
char s[maxn];
priority_queue<int,vector<int>,greater<int> >q;
int main()
{
    cin>>k>>s; sum=0;
    int n=strlen(s);
    for(int i=0;i<n;i++)
    {
        int t=s[i]-'0';
        sum+=t;
        q.push(t);
    }
    ans=0;
    while(sum<k)
    {
        int t=q.top();
        q.pop();
        sum+=(9-t);
        q.push(9);
        ans++;
    }
    cout<<ans<<endl;
}