【CodeForces 282A】Bit++（水~）

[align=center]A. Bit++[/align][align=center][/align][align=center][/align][align=center]time limit per test[/align][align=center]1 second[/align][align=center][/align][align=center][/align][align=center]memory limit per test[/align][align=center]256 megabytes[/align][align=center][/align][align=center][/align][align=center]input[/align][align=center]standard input[/align][align=center][/align][align=center][/align][align=center]output[/align][align=center]standard output[/align]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, calledx. Also, there are two operations:Operation ++ increases the value of variablex by 1.Operation -- decreases the value of variablex by 1.A statement in language Bit++ is a sequence, consisting of exactly one operation and one variablex. The statement is written without spaces, that is, it can only contain characters "+", "-", "X". Executinga 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 ofx is 0. Execute the programme and find its final value (the value of the variable when this programme is executed).InputThe 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 variablex (denoted as letter «X»). Thus, there are no empty statements. The operation and the variable can be written in any order.OutputPrint a single integer — the final value of x.ExamplesInput

1
++X

Output

1

Input

2
X++
--X

Output

0

题目大意：自增自减

思路：这种大水题我还wa了一发，想抽死自己，想复杂了，以为最后一个如果是x++的话输出x自增之前的值，结果，hh

#include<iostream>#include<cstdio>using namespace std;int main(){int n,x=0;string s;cin>>n;while(n--){cin>>s;if(s[1]=='+') x++;else x--;}cout<<x<<endl;return 0;}