AK_HuangYC的博客

Problem DescriptionGiven a positive integer N, you should output the most right digit of N^N.InputThe input contains several test cases. The first line of the input is a single integer T

Define f(n) as the number of ways to perform n in format of the sum of some positive integers. For instance, when n=4, we have  4=1+1+1+1  4=1+1+2  4=1+2+1  4=2+1+1  4=1+3  4=2+2  4=3+1

斐波那契数列的定义如下：F(0) = 0F(1) = 1F(n) = F(n - 1) + F(n - 2) (n >= 2)(1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...)给出n，求F(n)，由于结果很大，输出F(n) % 1000000009的结果即可。Input

给出3个正整数A B C，求A^B Mod C。例如，3 5 8，3^5 Mod 8 = 3。Input3个正整数A B C，中间用空格分隔。(1 Output输出计算结果Input示例3 5 8Output示例3#include#include#includeusing namespace std;typ

DescriptionFermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but n

#include#includeusing namespace std;typedef long long ll;int prime[70000];void is_prime(){ for(int i=2;i<=70000;i++) { if(!prime[i]) { for(int j=2*i;j<=70000;j+=i) prime[j]=1; } }}

求：3^0 + 3^1 +...+ 3^(N) mod 1000000007Input输入一个数N(0 Output输出：计算结果Input示例3Output示例40关于逆元的一些知识点：若对于数字A,C 存在X，使A * X = 1 (mod C) ,那么称X为 A 对C的乘法逆元。逆元的

Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.More formally, there are 3n gn

Problem DescriptionA为一个方阵，则Tr A表示A的迹（就是主对角线上各项的和），现要求Tr(A^k)%9973。Input数据的第一行是一个T，表示有T组数据。每组数据的第一行有n(2 Output对应每组数据，输出Tr(A^k)%9973。Sample Input22 21 00 13 9999999

Problem Description由0和1组成的串中，不能表示为由几个相同的较小的串连接成的串，称为本原串，有多少个长为n（n答案mod2008.例如，100100不是本原串，因为他是由两个100组成，而1101是本原串。Input输入包括多个数据，每个数据一行，包括一个整数n，代表串的长度。Output对于每个测试数据，输出一行，代表有多少

Jzzhu has invented a kind of sequences, they meet the following property:You are given x and y, please calculate fn modulo1000000007 (109 + 7).InputThe first line contains two inte