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菲波纳数列,该怎么处理

热度:67   发布时间:2016-04-24 02:11:24.0
菲波纳数列
public class FeiBo {
public static void Xiang(int n) {
int sum = 1;
int a0 = 0;
int a1 = 1;
if (n < 2) {
System.out.println(a1);
} else {
int an = a(n - 1) + a(n - 2);
for (int i = 2; i <= n; i++) {
sum += ai;
System.out.println(sum);
}
}
}
public static void main(String[] args) {
Xiang(1);
}
};

ai 怎么定义?


------解决方案--------------------
好像是 a[i] 吧
Java code
    public static void main(String args[]){        Scanner sc = new Scanner(System.in);    int n = sc.nextInt();    int[] fb = new int[10000];    fb[0] = 1;    fb[1] = 1;    for(int i = 0; i < n; i++){        fb[i + 2] = fb[i] + fb[i + 1];    }             System.out.println(fb[n - 1]);    }//----------------------------------------------------    public static void main(String args[]){      Scanner sc = new Scanner(System.in);    int n = sc.nextInt();    System.out.println(fb(n - 1));    }    public static int fb(int n){    if(n == 1 || n == 0){        return 1;    }else{        return fb(n - 1) + fb(n - 2);    }    }
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