3 solutions

  • 0
    @ 2025-9-27 14:39:15
    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
    	int k,g,m;cin>>k>>g>>m;
    	int mug=0,glass=0;
    	for(int i=1;i<=k;i++){
    		if(glass==g) glass=0;
    		else if(mug==0)mug=m;
    		else{
    			if(mug+glass>g){mug=mug-g+glass;glass=g;
    			;}
    			else glass=glass+mug,mug=0;
    		}
    	//	cout<<glass<<" "<<mug<<endl; 
    	}
    	cout<<glass<<' '<<mug;
    	return 0;
    }
    
    
    • -1
      @ 2025-5-11 14:59:30
      #include<bits/stdc++.h>
      using namespace std;
      int main()
      {
      	int k,g,m;cin>>k>>g>>m;
      	int mug=0,glass=0;
      	for(int i=1;i<=k;i++){
      		if(glass==g)glass=0;
      		else if(mug==0)mug=m;
      		else{
      			if(mug+glass>g){mug=mug-g+glass;glass=g;
      			;}
      			else glass=glass+mug,mug=0;
      		}
      	//	cout<<glass<<" "<<mug<<endl; 
      	}
      	cout<<glass<<' '<<mug;
      	return 0;
      }
      
      • -1
        @ 2025-4-25 13:49:14
        #include <bits/stdc++.h>
        using namespace std;
        
        int main() 
        {
        	int k,g,m;
        	int x=0,y=0,z;
        	cin>>k>>g>>m;
        	for(int i=0;i<k;i++)
        	{
        		if (x==g)x=0; //第一种操作 
        		else if(y==0)y=m; //第二种操作 
        		else //第三种操作 
        		{
        			z=min(g-x,y); //取决于第一个杯子可以装的和第二个杯子有的最小值 
        			x+=z,y-=z;
        		}
        	}
        	cout<<x<<" "<<y<<endl;
        	return 0;
        }
        
        
        • 1

        Information

        ID
        2341
        Time
        1000ms
        Memory
        256MiB
        Difficulty
        1
        Tags
        (None)
        # Submissions
        84
        Accepted
        21
        Uploaded By