Problem——282A. Bit++——Codeforces

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#Problem——282A. Bit++——Codeforces #C++

本文介绍了一种特殊的编程语言Bit++，该语言仅包含一个变量x及两种操作：增加和减少x的值。文章详细解释了如何执行Bit++语言的程序，并通过示例展示了如何计算程序执行后x的最终值。

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The classic programming language of Bitland is Bit++. This language is so peculiar and complicated.

The language is that peculiar as it has exactly one variable, called x. Also, there are two operations:

Operation ++ increases the value of variable x by 1.
Operation – decreases the value of variable x by 1.
A statement in language Bit++ is a sequence, consisting of exactly one operation and one variable x. The statement is written without spaces, that is, it can only contain characters “+”, “-”, “X”. Executing a statement means applying the operation it contains.

A programme in Bit++ is a sequence of statements, each of them needs to be executed. Executing a programme means executing all the statements it contains.

You’re given a programme in language Bit++. The initial value of x is 0. Execute the programme and find its final value (the value of the variable when this programme is executed).

Input

The first line contains a single integer n (1 ≤ n ≤ 150) — the number of statements in the programme.

Next n lines contain a statement each. Each statement contains exactly one operation (++ or --) and exactly one variable x (denoted as letter «X»). Thus, there are no empty statements. The operation and the variable can be written in any order.

Output

Print a single integer — the final value of x.

Examples

input

1
++X

output

input

2
X++
–X

output

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<iomanip>
#include<cstring>
#include<string>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define ll long long
#define mes(x,y) memset(x,y,sizeof(x))
#define maxn 2147483648+30
using namespace std;
ll gar(ll a,ll b){//最大公约数 
return b==0?a:gar(b,a%b);
}
int main(){
	long n;
	while(cin>>n){
		string s;long x=0;
		while(n--){
			cin>>s;
			if(s[1]=='+'){
				if(s[2]=='+')x++;
				else if(s[0]=='+')++x;
			}
			if(s[1]=='-'){
				if(s[2]=='-')x--;
				else if(s[0]=='-')--x;
			}
		}
		cout<<x<<endl;
	}
	return 0;
}