# UVA 10931 Parity
## 題目連結 [UVA 10931](https://vjudge.net/problem/UVA-10931)
### 題目內容
We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 101012 has three 1s in its binary representation so it has parity 3(mod2), or 1. In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647.
### 輸入限制
Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed.
### 輸出限制
For each integer I in the inputt you should print a line ‘The parity of B is P (mod 2).’, where B is the binary representation of I.
### 解題思路
這題要輸出二進制的數字，當temp除2的餘數等於0時push 0進字串中，反之則push 1進字串中，之後再將字串reverse就是二進制數字。
### 程式碼

#include<bits/stdc++.h>
using namespace std;
int main(){
    long long n;
    while(cin>>n && n!=0){
        string a;
        long long cot=0;
        while(n!=0){
            if(n%2==0){
                a.push_back('0');
            }
            else{
                a.push_back('1');
                cot++;
            }
            n/=2;
        }
        reverse(a.begin(),a.end());
        cout<<"The parity of "<<a<<" is "<<cot<<" (mod 2)."<<endl;
    }
}

## 測資
### Sample input
1
2
10
21
0
### Sample output
The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2).
## 中文題目連結 [zerojudge a132](https://zerojudge.tw/ShowProblem?problemid=a132)