1 条题解

  • 0
    @ 2025-4-7 21:19:28

    C :

    #include<stdio.h>
    int main()
    {
        int a[200];
        int n,i,x;
        while(scanf("%d",&n)!=EOF)
        {
        for(i=0;i<n;i++)
        scanf("%d",&a[i]);
        scanf("%d",&x);
        for(i=0;i<n;i++)
        {
            if(a[i]==x)
            {
                printf("%d\n",i);
                break;
            }
        }
        if(i==n)
        printf("-1\n");
        }
    }
    
    

    C++ :

    #include<stdio.h>
    
    int main()
    {
    	int n,x,i,a[200];
    	while(scanf("%d",&n)!=EOF)
    	{
    		for(i=0;i<n;i++)
    			scanf("%d",&a[i]);
    		scanf("%d",&x);
    		for(i=0;i<n;i++)
    			if(a[i]==x)
    			{
    				printf("%d\n",i);
    				break;
    			}
    		if(i==n)
    			puts("-1");
    	} 
    	return 0;
    }
    

    Pascal :

    var
       n,i,x:longint;
       a:array[0..1000]of longint;
       p:boolean;
    begin
     while not(eof(input))do
      begin
       readln(n);
       for i:=1 to n do read(a[i]);
       readln(x);
       p:=true;
       for i:=1 to n do
        if a[i]=x then begin writeln(i-1);p:=false;break;end;
       if p then writeln(-1);
      end;
    end.
    

    Java :

    
    import java.util.Scanner;
    
    public class Main {
    
    	public static void main(String[] args) {
    		// TODO Auto-generated method stub
    
    		Scanner in = new Scanner(System.in);
    		while (in.hasNext()) {
    			int n = in.nextInt();
    			int i = 0;
    			int num = 0;
    			int d = -1;
    
    			if (n <= 200 && n > 0) {
    				int[] a = new int[n];
    				for (int c = 0; c < a.length; c++)
    					a[c] = in.nextInt();
    
    				int x = in.nextInt();
    
    				for (i = 0; i < a.length; i++) {
    					if (x == a[i]) {
    						num++;
    						break;
    					}
    				}
    			}
    			if (num > 0)
    				System.out.println(i);
    			else
    				System.out.println(d);
    		}
    	}
    }
    
    
    • 1

    信息

    ID
    1122
    时间
    1000ms
    内存
    32MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者