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 <iostream> using namespace std; int main(){ int x,y,j,t; cin>>x>>y; j=(x*4-y)/2; t=x-j; cout<<j<<endl<<t; return 0; }

#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 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; }

#include<bits/stdc++.h> using namespace std; int main(){ int a,b; cin>>a>>b; for (int i=1;i<=a;i++){ if (i*2+(a-i)*4==b){ cout<<i<<endl<<a-i; break; } } return 0; }

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

鸡兔同笼逻辑分析（不会鸡兔同笼的同学有福啦！！！） #include <bits/stdc++.h> using namespace std; int main(){ int n,y; cin>>n>>y; int jiao =n*2; int duo=y-jiao; int tu=duo/2; int ji=n-tu; cout<<ji<<endl<<tu; return 0; }