A. Wrong Subtraction（模拟题，取余10获得最后一位）-优快云博客

本文介绍了一个特殊的减法算法，该算法由小女孩Tanya使用，当遇到数字末尾为零时，不是简单减一而是直接去除末位数字。文章通过一个具体的C语言实现示例，展示了如何根据这一规则多次减少给定数字。

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A. Wrong Subtraction
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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
n. Tanya will subtract one from it
k
k times. Your task is to print the result after all
k
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
n and
k
k (
2
≤
n
≤
10
9
2≤n≤109,
1
≤
k
≤
50
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
n by one
k
k times.

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

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

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int n,k,a;
    scanf("%d%d",&n,&k);
    int i;
    for(i=1;i<=k;i++)
    {
        a=n%10;//不管多大只要是取余就可以获取最后一位
        if(a==0)
        {
            n=n/10;
        }
        else n=n-1;
    }
    printf("%d\n",n);
    return 0;
}