#include<iostream> using namespace std; int main(){ int x,y,a,b; cin >> x >> y; a = y / 2 - x; b = x - a; cout << b; cout << endl; cout << a; return 0; }

#include<bits/stdc++.h> using namespace std; int main(){ int x,y; cin>>x>>y; int ji1=4*x-y; int ji2=4-2; int ji=ji1 / ji2; int tu=x-ji; cout<<ji<<endl<<tu; return 0; }

#include<bits/stdc++.h> using namespace std; int main(){ int a,b; cin>>a>>b; cout<<a-(b/2-a)<<endl<<b/2-a; return 0; }

空降: CP002966 代码如下: 此题可结合数学中鸡兔同笼的公式来解

#include <bits/stdc++.h> using namespace std; int cnt=0,n,m; int main(){ cin>>n>>m; for(int j=0;j<=n;++j){ for(int t=0;t<=n;++t){ if(j+tn and 2j+4tm){ cout<<j<<endl; cout<<t; } } } }

#include <iostream> using namespace std; int main(){ int n, m; cin >> n >> m; int tuzi = (m - n * 2) / 2; cout << n - tuzi << endl << tuzi; }

#include<bits/stdc++.h> using namespace std; int x,y,j,t; int main(){ cin>>x>>y; t=y/2-x; j=x-t; cout<<j<<endl; cout<<t<<endl; }

#include <iostream> using namespace std; int main() { int x,y; cin>>x>>y; for(int ji=1;ji<=x;ji++) { for(int tu=1;tu<=x;tu++) { if(tu+jix&&tu4+ji2y) { cout<<ji<<endl<<tu; break; } } } return 0; }

#include <iostream> using namespace std; int main(){ int x; int y; cin >> x >> y; cout << (4x-y)/2 <<endl <<(y-2x)/2; return 0; }

#include <iostream> using namespace std; int main() { int heads, legs; cin >> heads >> legs; }

#include<iostream> using namespace std; int main(){ int a,b,n; cin>>a>>b; n=b/2-a; cout<<a-n<<endl; cout<<n; return 0; }