两题水题: 1.如果一个数能被分解为两个素数的乘积,则称为Semi-Prime,给你一个数,让你判断是不是Semi-Prime数. 2.定义F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2) 让你判断第n项是否能被3整除. 1.ZOJ 2723 Semi-Prime http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1723 打表即可. #include<cstdio>…

题目链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2060 题目描述: 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. (…

https://vjudge.net/contest/67836#problem/A 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 wo…

Power of Fibonacci Time Limit: 5 Seconds      Memory Limit: 65536 KB In mathematics, Fibonacci numbers or Fibonacci series or Fibonacci sequence are the numbers of the following integer sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, .…

A Fibonacci sequence is calculated by adding the previous two members of the sequence, with the first two members being both 1. f(1) = 1, f(2) = 1, f(n > 2) = f(n - 1) + f(n - 2) Your task is to take a number as input, and print that Fibonacci number…

In mathematics, Fibonacci numbers or Fibonacci series or Fibonacci sequence are the numbers of the following integer sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ... By definition, the first two numbers in the Fibonacci sequence are…

题意:在给定的数组里,寻找一个最长的序列,满足ai-2+ai-1=ai.并输出这个序列. 很容易想到一个DP方程 dp[i][j]=max(dp[k][i])+1. (a[k]+a[i]==a[j],1<=k&&k<i) dp[i][j]表示序列最后两位是a[i],a[j]时的最长长度. 这个方程状态是O(n^2),转移是O(n),总复杂度是O(n^3)会超时. 进一步思考会发现转移这里是可以优化的.实际上我们只需要知道离i最近的那个满足a[k]+a[i]==a[j]的k就行,…

题目 //f(1) = 1, f(2) = 1, f(n > 2) = f(n - 1) + f(n - 2) import java.io.*; import java.util.*; import java.math.*; public class Main { /** * @xqq */ public BigInteger an(int n) { BigInteger c; BigInteger a = BigInteger.valueOf(1); BigInteger b = BigIn…

解题思路: 找规律,不难的,打表 坑的地方在于题目限定条件 and the seed value for G(1) is a random integer t, (t>=1) 虽然都用粗体表示出来了= = 但我还是没注意到 = = #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main(){ int n…

题目大意: 求斐波那契数列前n项的k次幂和  Mod 1000000009.    n<=1e18, k<=1e5 这题的k比较大,所以不能用矩阵乘法来递推.学到了新姿势...  http://blog.csdn.net/acdreamers/article/details/23039571 基本思想就是求出通项公式,把里面的$\sqrt{5}$   用 $x$ 代替, 其中 $x^2\equiv 5\pmod{1000000009}$ 然后二项式展开求和就好了. 一个合法的$x$是38300…