1 solutions

  • 0
    @ 2023-12-3 21:35:37

    C :

    #include<stdio.h>
    int main()
    {
    	int n,i,k;
    	char a[200],b,c;
    
    #ifndef ONLINE_JUDGE
    	freopen("in.txt","r",stdin);
    #endif
    	scanf("%d\n",&n);
    	while(n--)
    	{
    		i=0;k=0;
    		gets(a);
    		for(i=0;a[i]!='\0';i++)
    		{
    			if(a[i]==' ')
    				k=0;
    			else
    			{
    				if(k==0)
    				{
    					if(a[i]>='a')
    						a[i]-=32;
    					printf("%c",a[i]);
    					k=1;
    				}
    				else 
    				{
    					k=1;
    				}
    			}
    		}
    		printf("\n");
    	}
    return 0;
    }
    

    C++ :

    /*
    这一题很简单,用来练习字符串处理函数正好。
    */
    #include<stdio.h>
    #include<string.h>
    #include<ctype.h>
    
    int main()
    {
        int t;
        char c[101];
        int k,i;
        char * temp[10];
        scanf("%d\n",&t);
        for(k=0;k<t;k++)
        {
            gets(c);
            temp[0]=strtok(c," ");
            printf("%c",toupper(temp[0][0]));
            for(i=1;i<10;i++)
            {
                temp[i]=strtok(NULL," ");
                if(temp[i]==NULL)break;
                printf("%c",toupper(temp[i][0]));
            }
            printf("\n");
        }
        return 0;
    }
    

    Java :

    import java.util.Scanner;
    import java.util.StringTokenizer;
    
    public class Main{
    	public static void main(String args[]){
    		Scanner in = new Scanner(System.in);
    		int T = in.nextInt();
    		in.nextLine();
    		while(T-- != 0){
    			StringTokenizer s = new StringTokenizer(in.nextLine(), " ");
    			while(s.hasMoreTokens()){
    				System.out.print(((s.nextToken()).toUpperCase()).charAt(0));
    			}
    			System.out.println();
    		}
    	}
    }
    
    • 1

    Information

    ID
    1325
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    # Submissions
    0
    Accepted
    0
    Uploaded By