zoukankan      html  css  js  c++  java
  • CF w1d1 A. A pile of stones

    A. A pile of stones

    Vasya has a pile, that consists of some number of stones. n times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.

    You are given n operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.

    Input

    The first line contains one positive integer n — the number of operations, that have been made by Vasya (1≤n≤100).

    The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, si is equal to "-" (without quotes), if added, si is equal to "+" (without quotes).

    Output

    Print one integer — the minimal possible number of stones that can be in the pile after these n operations.

    Note

    In the first test, if Vasya had 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).

    In the second test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.

    In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. It can be proved, that it is impossible to have less number of stones after making the operations.

    In the fourth test, if Vasya had 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3.

    #include<iostream>
    #include<string>
    using namespace std;
    int main()
    {
    	int n,mark,ans=0;
    	string s;
    	cin>>n;
    	cin>>s;
    	if(s[0]=='+'){
    		for(int i=0;i<s.size();i++){
    			if(s[i]=='+')ans++;
    			if(s[i]=='-'&&ans>=1)ans--;
    		}
    	}
    	else
    	{
    		for(int i=0;i<s.size();i++)if(s[i]!='-'){
    			mark=i;
    			break;
    		}
    		for(int i=mark;i<s.size();i++){
    			if(s[i]=='+')ans++;
    			if(s[i]=='-'&&ans>=1)ans--;
    		}
    	}
    	cout<<ans;
    	return 0;
    } 
    
  • 相关阅读:
    如何在Unity中播放影片
    C# typeof()实例详解
    unity3d用鼠标拖动物体的一段代码
    unity3d中Find的用法
    geometry_msgs/PoseStamped 类型的变量的构造
    c++ ros 计算两点距离
    C++ 利用指针和数组以及指针和结构体实现一个函数返回多个值
    C++ 结构体指针的定义
    Cannot initialize a variable of type 'Stu *' with an rvalue of type 'void *'
    C++中的平方、开方、绝对值怎么计算
  • 原文地址:https://www.cnblogs.com/LiangYC1021/p/12656062.html
Copyright © 2011-2022 走看看