oj练习1

题目（难度***）

Desc
Calculate expression.

Input
The input contains several test cases, one test case per line.
Each case have nintegers(0<=n<2
31
) and operators(“+” or “-”), separated by blank spaces.The number of integers is small than 10.

Output
One case per line, output the answer

输入样例

123456789 + 8 - 765
5 - 987456 + 2

输出样例

123456032
-987449

收获

1：学会了多组string输入，参考链接

代码

#include <iostream>
#include <bits/stdc++.h>

using namespace std;


int main()
{
    string s;
    while(getline(cin,s)){
    int num=0;
    queue<char >op;
    queue<int >number;
    for(int i=0;i<s.length();i++){
        if(s[i]=='+'||s[i]=='-')
          op.push(s[i]);
        else if(s[i]==' ')
                continue;
        else{
            while(s[i]!=' '&&i<s.length())
            {
                num=num*10+s[i]-'0';
                i++;
            }
            //cout<<num<<endl;
                number.push(num);
                num=0;
        }
    }
    long long sum=number.front();
    number.pop();
    while(!number.empty()){
        int temp=number.front();
        number.pop();
        int oper=op.front();
        op.pop();
        if(oper=='+')
            sum+=temp;
        else{
            sum-=temp;
        }
    }
    cout<<sum<<endl;
    }
    return 0;
}