1352B - Same Parity Summands - CodeForces Solution


constructive algorithms math *1200

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Python Code:

t = int (input ());

for i in range (t) :
    n, k = map (int, input ().split ());
    tmp1 = n - 2 * (k - 1);
    tmp2 = n - (k - 1);
    if ((tmp1 > 0) and (tmp1 % 2 == 0)) :
        print ("YES");
        print ("2 " * (k - 1), tmp1, sep='');
    elif ((tmp2 > 0) and (tmp2 % 2 == 1)) :
        print ("YES");
        print ("1 " * (k - 1), tmp2, sep='');
    else :
        print ("NO");

C++ Code:

#include<bits/stdc++.h>
using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);


    int t;
    cin>>t;
    while(t--)
    {
        int n,k,m;
        cin>>n>>k;
        m=n;
            for(int i=1; i<k; i++)
            {
                n=n-2;
                m=m-1;
            }
            if((n%2!=0||n<2)&&(m%2==0||m<1))
            cout<<"NO"<<endl;
            else if(n%2==0&&n>=2)
            {
                cout<<"YES"<<endl<<n<<" ";
                for(int j=1; j<k; j++)
                cout<<"2"<<" ";
                cout<<endl;
            }
            else if(m%2!=0&&m>=1)
            {
                cout<<"YES"<<endl<<m<<" ";
                for(int j=1; j<k; j++)
                cout<<"1"<<" ";
                cout<<endl;
            }
    }

    return 0;
}


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