UVa problem solution 10931 - Parity

problem link:


Discuss: in this problem you just simply convert decimal to binary. then count the number of one's in binary that will be the 'P'. and for converting you can use c++ built in STL     string binary = std::bitset<31>(n).to_string();   . and there is no tricky part in this problem.

try yourself, before see the code




#include<bits/stdc++.h>
using namespace std;
void parity(long long n)
{
    long long one=0,m,i,flag=0;
    string binary = std::bitset<31>(n).to_string();
    m=binary.length();
    cout <<"The parity of ";
    for(i=0;i<m;i++)
    {
        if(binary[i]=='1')
            flag=1;
        if(flag)
            cout <<binary[i];

    }
    cout <<" is ";
        sort(binary.begin(),binary.end());

    for(i=0; i<m; i++)
    {
        if(binary[i]=='1')
            one++;
    }
    printf("%lld (mod 2).\n",one);

}
int  main()
{
    long long n,i,j;
    while(cin >>n&&n)
    {

        parity(n);
    }
    return 0;

}