Fibonacci数列C语言三种实现方法 技术标签: c语言 算法 其他Fibonacci数列的数学公式 列举:1,1, 2, 3, 5, 8, 13、、、第三项等于第一项与第二项的和运用数组求解#include <stdio.h> int main (void) { int i, i_num, F_num[10000]; F_num[1] = F_num[2] = 1;//下标为1时,时斐波那契而...
Discover What is Fibonacci series in C, a technique that involves calling a function within itself to solve the problem. Even know how to implement using different methods.
【C语言】fibonacci数列 技术标签: C语言1、输出fibonacci数列前N位 void main() { unsigned long num1 = 1; unsigned long num2 = 1; unsigned long tmp = 0; int i = 0; int n=1; printf("请输入n:"); scanf("%d",&n); if (n < 3) { printf("请输入一个大于等于3的数!\n"); ...
every number in the sequence is regarded as a term denoted by the expression Fn. The n indicates where the given number falls in the sequence, which starts at 0. For instance, the fourth term is known as F3, and the eighth term is known as F7. ...
代码语言:javascript 代码运行次数:0 运行 AI代码解释 deffib(n,memorize={1:0,2:1}):ifninmemorize:returnmemorize[n]memorize[n]=fib(n-1,memorize)+fib(n-2,memorize)returnmemorize[n] 时间复杂度为O(n), 空间复杂度为O(n) 3. 递归+两变量 ...
因为题目保证pmod10pmod10是一个完全平方数,也就是说pmod5pmod5等于1或-1,即5是模pp的二次剩余(据说)。 求出通项,用Cipolla求出5的二次剩余,记为cc,并记p=1+c2p=1+c2, 通项变成 解得 然后枚举一下nn的奇偶性,再用BSGS求出nn就可以了。
importjava.io.*;importjava.util.*;importjava.math.*;publicclassFibonacci{// Returns n-th Fibonacci numberstaticBigIntegerfib(int n){BigInteger a=BigInteger.valueOf(0);BigInteger b=BigInteger.valueOf(1);BigInteger c=BigInteger.valueOf(1);for(int j=2;j<=n;j++){c=a.add(b);a=b;b=c;...
A high-level description of what it does is: store two last numbers in variables c4 and c5 (initially c4=0, c5=1), print the number stored in c5 (this operation takes the major part of the code), calculate next number (c6 = c5+c4), and move the numbers sequence one number back ...
Codechef:Fibonacci Number/FN——求通项+二次剩余+bsgs 题意 定义$F_n$ 为 $$F_n = \left\{\begin{matrix} 0, n=0\\ 1, n=1 \\ F_{n-1} + F_{n-2}, n > 1 \end{matrix}\right.$$ 现给你一个素数 $p$ 和一个非负整数 $C$,你需要最小的非负整数 $n$,使得 $F_n \equiv...
【CF446C】DZY Loves Fibonacci Numbers(线段树) Description 给定一个序列,资瓷区间加上一个斐波那契数列,区间求和。 Solution 有一个性质: fib[a+b]=fib[a−1]×fib[b]+fib[a]×fib[b+1] f i b [ a + b ] = f i b [ a − 1 ] × f i b [ b ] + f i b [ a ] × f i ...