1 solutions

  • 0
    @ 2025-11-5 17:32:13

    C :

    #include <stdio.h>
    
    int a[1001] = {0,1,1};
    
    int main(void) {
    	int i, n;
    	do
    		scanf("%d", &n);
    	while(!(n >= 1 && n <= 1000));
    	for(i = 3; i <= n; i ++)
    		a[i] = a[i-2] + a[i-1];
    	printf("%d", a[n]);
    	return 0;
    }
    
    

    C++ :

    #include<cstdio>
    using namespace std;
    int a[110]; 
    int main()
    {
    	int i,n;
    	scanf("%d",&n);
    	a[1]=1;
    	a[2]=1;
    	for(i=3;i<=n;i++)
    	{
    		a[i]=a[i-1]+a[i-2];
    	}
    	printf("%d\n",a[n]);
    	return 0;
    }
    

    Pascal :

    var i,j,k,m,n:longint;
      a:array[1..10000]of longint;
    begin
      readln(n);
      a[1]:=1;
      a[2]:=1;
      for i:=3 to n do
        a[i]:=a[i-1]+a[i-2];
      write(a[n]);
    end.
    
    • 1

    第四章:for循环结构《练习6:Fibonacci序列》

    Information

    ID
    18533
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    # Submissions
    0
    Accepted
    0
    Uploaded By