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注意这题要考虑 (s-d)是奇数的情况就可以了…………#include<iostream> using namespace std; int main() { int n; int sum; int mul; cin>>n; for(int i =0; i<n; i++) { cin>>sum>>mul; if(sum<mul) cout<<"impossible"<<endl; else if((sum-mul)%2==0) cout<<(sum+mul)/2<<" "<<(sum-mul)/2<<endl; else cout<<"impossible"<<endl;//如果s-d是奇数也是impossible…… } return 0; } Followed by:
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