Fibonacci公式的递推的步骤的过程的实现，怎么利用C语言的程序的设计的思维来实现的呢？

Problem Description
There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).

Input
Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).

Output
Print the word "yes" if 3 divide evenly into F(n).

Print the word "no" if not.

Sample Input
0
1
2
3
4
5

Sample Output
no
no
yes
no
no
no

Fibonacci数列的递推公式为：Fn=Fn-1+Fn-2，其中F1=F2=1。 当n比较大时，Fn也非常大，现在我们想知道，Fn除以10007的余数是多少。 public void solve2(int a){ int [] arr=new int[a+1]; arr[1]=arr[2]=1; for(int i=3;i<arr.length;i++){ arr[i]=(arr[i-1]+arr[i-2])%10007;//对每一个斐波那契数列进行取余数 } System.out.println(arr[a]); } 我想问一下 arr[i]=(arr[i-1]+arr[i-2])%10007 对每一斐波那契数列进行取余是什么意思 如何理解 我是这样理解的 例如 当i=4时 余数为 i=3的余数加i=2的余数 再除以10007取余 这样对吗 为什么等于先求值再取余
c++问题求解答！！！！
Fibonacci数列的递推公式为Fn=Fn-1 + Fn-2其中F1=F2=1 求Fn除以10007的余数是多少
Interesting Fibonacci 具体的实现
Problem Description In mathematics, the Fibonacci numbers are a sequence of numbers named after Leonardo of Pisa, known as Fibonacci (a contraction of filius Bonaccio, "son of Bonaccio"). Fibonacci's 1202 book Liber Abaci introduced the sequence to Western European mathematics, although the sequence had been previously described in Indian mathematics. The first number of the sequence is 0, the second number is 1, and each subsequent number is equal to the sum of the previous two numbers of the sequence itself, yielding the sequence 0, 1, 1, 2, 3, 5, 8, etc. In mathematical terms, it is defined by the following recurrence relation: That is, after two starting values, each number is the sum of the two preceding numbers. The first Fibonacci numbers (sequence A000045 in OEIS), also denoted as F[n]; F[n] can be calculate exactly by the following two expressions: A Fibonacci spiral created by drawing arcs connecting the opposite corners of squares in the Fibonacci tiling; this one uses squares of sizes 1, 1, 2, 3, 5, 8, 13, 21, and 34; So you can see how interesting the Fibonacci number is. Now AekdyCoin denote a function G(n) Now your task is quite easy, just help AekdyCoin to calculate the value of G (n) mod C Input The input consists of T test cases. The number of test cases (T is given in the first line of the input. Each test case begins with a line containing A, B, N, C (10<=A, B<2^64, 2<=N<2^64, 1<=C<=300) Output For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the value of G(N) mod C Sample Input 1 17 18446744073709551615 1998 139 Sample Output Case 1: 120

Problem Description Fibonacci数列，定义如下： f(1)=f(2)=1 f(n)=f(n-1)+f(n-2) n>=3。 计算第n项Fibonacci数值。 Input 输入第一行为一个整数N，接下来N行为整数Pi（1<=Pi<=1000）。 Output 输出为N行，每行为对应的f(Pi)。 Sample Input 5 1 2 3 4 5 Sample Output 1 1 2 3 5
Fibonacci的c++代码出现segmentation fault

Interesting Fibonacci 的编写
Problem Description In mathematics, the Fibonacci numbers are a sequence of numbers named after Leonardo of Pisa, known as Fibonacci (a contraction of filius Bonaccio, "son of Bonaccio"). Fibonacci's 1202 book Liber Abaci introduced the sequence to Western European mathematics, although the sequence had been previously described in Indian mathematics. The first number of the sequence is 0, the second number is 1, and each subsequent number is equal to the sum of the previous two numbers of the sequence itself, yielding the sequence 0, 1, 1, 2, 3, 5, 8, etc. In mathematical terms, it is defined by the following recurrence relation: That is, after two starting values, each number is the sum of the two preceding numbers. The first Fibonacci numbers (sequence A000045 in OEIS), also denoted as F[n]; F[n] can be calculate exactly by the following two expressions: A Fibonacci spiral created by drawing arcs connecting the opposite corners of squares in the Fibonacci tiling; this one uses squares of sizes 1, 1, 2, 3, 5, 8, 13, 21, and 34; So you can see how interesting the Fibonacci number is. Now AekdyCoin denote a function G(n) Now your task is quite easy, just help AekdyCoin to calculate the value of G (n) mod C Input The input consists of T test cases. The number of test cases (T is given in the first line of the input. Each test case begins with a line containing A, B, N, C (10<=A, B<2^64, 2<=N<2^64, 1<=C<=300) Output For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the value of G(N) mod C Sample Input 1 17 18446744073709551615 1998 139 Sample Output Case 1: 120
Interesting Fibonacci 的实现思维是什么
Problem Description In mathematics, the Fibonacci numbers are a sequence of numbers named after Leonardo of Pisa, known as Fibonacci (a contraction of filius Bonaccio, "son of Bonaccio"). Fibonacci's 1202 book Liber Abaci introduced the sequence to Western European mathematics, although the sequence had been previously described in Indian mathematics. The first number of the sequence is 0, the second number is 1, and each subsequent number is equal to the sum of the previous two numbers of the sequence itself, yielding the sequence 0, 1, 1, 2, 3, 5, 8, etc. In mathematical terms, it is defined by the following recurrence relation: That is, after two starting values, each number is the sum of the two preceding numbers. The first Fibonacci numbers (sequence A000045 in OEIS), also denoted as F[n]; F[n] can be calculate exactly by the following two expressions: A Fibonacci spiral created by drawing arcs connecting the opposite corners of squares in the Fibonacci tiling; this one uses squares of sizes 1, 1, 2, 3, 5, 8, 13, 21, and 34; So you can see how interesting the Fibonacci number is. Now AekdyCoin denote a function G(n) Now your task is quite easy, just help AekdyCoin to calculate the value of G (n) mod C Input The input consists of T test cases. The number of test cases (T is given in the first line of the input. Each test case begins with a line containing A, B, N, C (10<=A, B<2^64, 2<=N<2^64, 1<=C<=300) Output For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the value of G(N) mod C Sample Input 1 17 18446744073709551615 1998 139 Sample Output Case 1: 120

Problem Description Fibonacci数列，定义如下： f(1)=f(2)=1 f(n)=f(n-1)+f(n-2) n>=3。 计算第n项Fibonacci数值。 Input 输入第一行为一个整数N，接下来N行为整数Pi（1<=Pi<=1000）。 Output 输出为N行，每行为对应的f(Pi)。 Sample Input 5 1 2 3 4 5 Sample Output 1 1 2 3 5
M斐波那契数列 是怎么写的
Problem Description M斐波那契数列F[n]是一种整数数列，它的定义如下： F[0] = a F[1] = b F[n] = F[n-1] * F[n-2] ( n > 1 ) 现在给出a, b, n，你能求出F[n]的值吗？ Input 输入包含多组测试数据； 每组数据占一行，包含3个整数a, b, n（ 0 <= a, b, n <= 10^9 ） Output 对每组测试数据请输出一个整数F[n]，由于F[n]可能很大，你只需输出F[n]对1000000007取模后的值即可，每组数据输出一行。 Sample Input 0 1 0 6 10 2 Sample Output 0 60

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#include <iostream> #include <cstdlib> #include "MTime.h" using namespace std; long long _fibonacci(int i) { if (i == 0) return 0; if (i == 1) return 1; return _fibonacci(i-2) + _fibonacci(i-1); } int main() { ofstream out("fibonacci.txt"); int i = 3; long long maxNum = 0; long long num = 1; int useTime; while (maxNum < num) { MTime start; maxNum = num; num = _fibonacci(i++); MTime end; useTime = end.delta(&start); } out << "Fibonacci数列中" << endl << "能用long long数据类型存储的" << "最大数是第 " << i-1 << " 个数" << endl << "这个数是 " << maxNum << endl << "计算出这个数用时 " << useTime << " 秒" <<endl; cout << "Fibonacci数列中" << endl << "能用long long数据类型存储的" << "最大数是第 " << i-1 << " 个数" << endl << "这个数是 " << maxNum << endl << "计算出这个数用时 " << useTime << " 秒" <<endl; system("pause"); return 0; } 我算是一个C++初学者吧，上面程序是我自己写的。我运行它，一个晚上过去了还是没结果。貌似无限循环了。下班回来用F10测试，发现在“i==40”时还能正常进行，当运行到“i==41”时就一直等待了。我一再检查程序没发现问题出在哪儿。求大神解惑~~~ 另：我在将long long 类型用unsigned char类型替代时程序能正常运行且很快得出结果！

Problem Description 任何一个大学生对菲波那契数列(Fibonacci numbers)应该都不会陌生，它是这样定义的： F(1)=1; F(2)=2; F(n)=F(n-1)+F(n-2)(n>=3); 所以，1,2,3,5,8,13……就是菲波那契数列。 在HDOJ上有不少相关的题目，比如1005 Fibonacci again就是曾经的浙江省赛题。 今天，又一个关于Fibonacci的题目出现了，它是一个小游戏，定义如下： 1、 这是一个二人游戏; 2、 一共有3堆石子，数量分别是m, n, p个； 3、 两人轮流走; 4、 每走一步可以选择任意一堆石子，然后取走f个； 5、 f只能是菲波那契数列中的元素（即每次只能取1，2，3，5，8…等数量）； 6、 最先取光所有石子的人为胜者； 假设双方都使用最优策略，请判断先手的人会赢还是后手的人会赢。 Input 输入数据包含多个测试用例，每个测试用例占一行，包含3个整数m,n,p（1<=m,n,p<=1000）。 m=n=p=0则表示输入结束。 Output 如果先手的人能赢，请输出“Fibo”，否则请输出“Nacci”，每个实例的输出占一行。 Sample Input 1 1 1 1 4 1 0 0 0 Sample Output Fibo Nacci
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Fibonacci Again的一个编程实现的做法
Problem Description There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2). Input Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000). Output Print the word "yes" if 3 divide evenly into F(n). Print the word "no" if not. Sample Input 0 1 2 3 4 5 Sample Output no no yes no no no

var a[21]; a[0]=0; a[1]=1; for(var i=2;i<=20;i++) a[i]=a[i-1]+a[i-2]; for(var j=1;j<21;j++) { document.write(a[j]); } 这样为什么无法实现斐波那契中的前20项，而在C语言中可以实现

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