1 solutions

  • 0
    @ 2025-11-5 19:47:03

    C :

    #include <stdio.h>
    int fib(int x);
    int main(void)
    {
    	int sum,m,n,i,x;
    	scanf("%d%d",&m,&n);
    	for (i=1;;i++)
    	{
    		x=fib(i);
    		if(x>=m&&x<=n)
    		sum+=x;
    		else if(x>n)
    		break;
    	}
    	printf ("%d",sum);
    	return 0;
    }
    int fib(int x)
    {
    	int a[3],i,j,sum;
    	a[0]=1,a[1]=1,a[2]=2;
    	if (x==1)
    	return 1;
    	if (x==2)
    	return 1;
    	if (x==3)
    	return 2;
    	for (i=4;i<=x;i++)
    	{
    		a[0]=a[1];
    		a[1]=a[2];
    		a[2]=a[0]+a[1];	
    	}
    	return a[2];
    }
    

    C++ :

    #include<iostream>
    using namespace std;
    int fib(int n)	//递归
    {
    	if (n==1 || n==2) return 1;
    	int f1=1,f2=1,f;
    	for (int i=3; i<=n; i++)
    		{ f=f1; f1=f2; f2+=f; }
    	return f2;
    }
    int main()
    {
    	int m,n,i=1,t,sum=0;
    	cin>>m>>n;
    	do
    	{
    		t=fib(i++);
    		if (t>=m && t<=n) sum+=t;
    	} while (t<n);
    	cout<<sum<<endl;
    	return 0;
    }
    
    • 1

    《C语言程序设计》江宝钏主编-习题6-8-斐波那契部分和

    Information

    ID
    19665
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
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