Codeforces Round #427 (Div. 2) B. The number on the board

最新推荐文章于 2024-07-27 23:44:56 发布

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最新推荐文章于 2024-07-27 23:44:56 发布

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本文介绍了一道算法题目，给定一个数k和一串自然数，目标是最少改变多少位数使得自然数总和大于等于k。文章提供了解决方案及AC代码。

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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.

这道题题意就是先给你一个数k，然后给你一串自然数。求最少多少位自然数经过变化可以使整串自然数相加的和大于等于k。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
char k[1111111];
int cnt[11];
int main()
{
    long long int n;
    while(cin>>n)
    {
        scanf("%s",k);
        memset(cnt,0,sizeof(cnt));
        for(int i=0;k[i];i++)
        {
            n-=(k[i]-'0');
            cnt[k[i]-'0']++;
            if(n<=0)
                break;
        }
        if(n<=0)
        {
            cout<<0<<endl;
            continue;
        }
        int res=0;
        for(int i=0;i<=9;i++)
        {
            while(cnt[i])
            {
                n-=(9-i);
                res++;
                cnt[i]--;
                if(n<=0)
                    break;
            }
            if(n<=0)
                break;
        }
        cout<<res<<endl;
    }
    return 0;
}