1 条题解

  • 0
    @ 2025-10-10 15:45:56

    C :

    #include<stdio.h>
    int main(){
    	int n,i,x,y;
    	scanf("%d",&n);
    	int a[n+1];
    	for(i=0;i<n;i++){
    		scanf("%d",&a[i]);
    	}
    	scanf("%d%d",&x,&y);
    	for(i=n-1;i>=x-1;i--){
    		a[i+1] = a[i];
    	}
    	a[x-1] = y;
    	for(i=0;i<n+1;i++){
    		printf("%d ",a[i]);
    	}
    	return 0;
    }
    
    

    C++ :

    #include <iostream>
     
    using namespace std;
     
    int main(){
        int n,i,t,x,y;
        cin>>n;
        int a[n + 1];
        for (i = 0;i < n;i++){
            cin>>a[i];
        }
        cin>>x;
        x--;
        cin>>y;
        for (i = n - 1;i >= x;i--){
            a[i + 1] = a[i];
        }
        a[x] = y;
        for (i = 0;i < n + 1;i++){
            cout<<a[i]<<" ";
        }
    }
    

    Pascal :

    var
    a:array[1..1000]of longint;
    n,i,x,y:longint;
    begin
    readln(n);
    for i:=1 to n do
    begin
    read(a[i]);
    end;
    readln;
    readln(x);
    readln(y);
    for i:=1 to x-1 do
    write(a[i],' ');
    write(y,' ');
    for i:=x to n-1 do
    write(a[i],' ');
    writeln(a[n]);
    readln;
    end.
    

    Java :

    import java.util.Scanner;
    
    public class Main{
    	public static void main(String[] args){	
    
        Scanner sc = new Scanner(System.in);
        int n=sc.nextInt();
        int a[] = new int[n+1];
        for(int i=0;i<a.length-1;i++) {
        a[i] = sc.nextInt();
        }
        //待插入的下标
        int j = sc.nextInt()-1;
        for(int i= a.length-2;i>=0;i--) {
        if(i>=j) {
        a[i+1] = a[i];
        }
        }
        a[j] = sc.nextInt();
        for(int i=0 ; i<a.length ;i++) {
        System.out.println(a[i]);
        } 
      
    }
    }
    

    Python :

    n=int(input())
    l1=list(map(int,input().split()))
    x=int(input())
    y=int(input())
    l1.insert(x-1,y)
    for i in l1:
        print(i,end=' ')
    
    • 1

    信息

    ID
    206
    时间
    1000ms
    内存
    256MiB
    难度
    10
    标签
    递交数
    2
    已通过
    1
    上传者