摘要: Solution with loop:#include using namespace std;int main(){ int m,n,k,r; while(std::cin>>m>>n>>k){ r=n-m; coutusing namespace std;void operation(int m,int n,int mnow,int nnow,int k){ if(nnow==k) return; if(nnow>(m-mnow)){ nnow=nnow-(m-mnow); cout>m>>n>>k){ mnow=0; 阅读全文
posted @ 2014-01-06 16:54 丸子No1 阅读(274) 评论(0) 推荐(0) 编辑