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.
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.
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.
512 4
50
1000000000 9
1
The first example corresponds to the following sequence: 512→511→510→51→50.
题意:CF第一套div3,难度开始下降,1600分以上的做了不会算rating,第一次是测试的,注册的人很多,很多大佬都AK了,仰望.....输入一个数n,和k,对这个数进行k次操作,每次可以有两种操作方式,如果这个数的个数为0,就可以/10,相当于把0去掉,如果不为0,就减一。经过k次操作,输出当前n的值。
题解:模拟
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#include<bits/stdc++.h> using namespace std; int main() { int n,k; cin>>n>>k; while(k--) { if(n%10==0) n/=10; else n--; } cout<<n; return 0; }