Codeforces Round #479 (Div. 3)

最新推荐文章于 2025-07-02 17:27:16 发布

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最新推荐文章于 2025-07-02 17:27:16 发布

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本文介绍了一个有趣的编程挑战，名为纠正错误减法，Tanya使用了一种独特的减法规则来减少数字。如果数字的最后一位是非零数，则减一；如果是零，则除以十。任务是在进行k次操作后找到最终的数字。

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A. Wrong Subtraction
题目链接

Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:

if the last digit of the number is non-zero, she decreases the number by one;
if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number n. Tanya will subtract one from it k times. Your task is to print the result after all k subtractions.

It is guaranteed that the result will be positive integer number.

Input
The first line of the input contains two integer numbers n and k (2≤n≤109, 1≤k≤50) — the number from which Tanya will subtract and the number of subtractions correspondingly.

Output
Print one integer number — the result of the decreasing n by one k times.

It is guaranteed that the result will be positive integer number.

Examples
input
512 4
output
50
input
1000000000 9
output
1
Note
The first example corresponds to the following sequence: 512→511→510→51→50.

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
const int maxn=20000+100;
typedef long long ll;
int main()
{
    int n,k;
    int sum;
    cin>>n>>k;
    for(int i=0;i<k;i++)
    {
        if(n%10==0)
        {
            n/=10;
        }
        else
        {
            n-=1;
        }
    }
    cout<<n;
    return 0;
}