Description Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,C<=1000000000,1<=B<=10^1000000). Input There are multiply testcases. Each testcase, there is one line contains three integers A, B and C, separated by a sin
Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 3245 Accepted Submission(s): 1332 Problem Description Sample Input 2 Sample Output 2 Hint 1. For N = 2, S(1) = S(2) = 1. 2. The input
题目链接:https://www.nowcoder.com/acm/contest/142/A 题目描述 A ternary string is a sequence of digits, where each digit is either 0, 1, or 2. Chiaki has a ternary string s which can self-reproduce. Every second, a digit 0 is inserted after every 1 in the str
Description Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,C<=1000000000,1<=B<=10^1000000). Input There are multiply testcases. Each testcase, there is one line contains three integers A, B and C, separated by a singl
Problem 1759 Super A^B mod C Accept: 1056 Submit: 3444Time Limit: 1000 mSec Memory Limit : 32768 KB Problem Description Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,C<=1000000000,1<=B<=10^1000000). Input There
传送门:http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=3151 时间限制(普通/Java):1000MS/3000MS 内存限制:65536KByte 描述 H1N1 like to solve acm problems.But they are very busy, one day they meet a problem. Given three intergers a,b,c, the
题目链接:http://acm.fzu.edu.cn/problem.php?pid=1759 欧拉降幂是用来干啥的?例如一个问题AB mod c,当B特别大的时候int或者longlong装不下的时候需要对B mod 一个数并且不改变最终的答案,这就可以将使用欧拉降幂公式来进行计算. 公式:AB mod C == AB%φ(C) + φ(C) mod C (其实φ(c)是c的欧拉函数值),这样就是欧拉函数的使用. #include <stdio.h> #include <cstrin
https://nanti.jisuanke.com/t/41299 题意:让算a^(a^(a^(...))),一共b个a, (mod p)的结果. 思路:这是个幂塔函数,用欧拉降幂公式递归求解. #include<bits/stdc++.h> #define ll long long using namespace std; map<int,int> euler; ll a,b,mod; int phi(int n) { int now=n; int ret=n; if(eule