Raising Modulo Numbers Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 5532 Accepted: 3210 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…
Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that…
ZOJ2150 快速幂,但是用递归式的好像会栈溢出. #include<cstdio> #include<cstdlib> #include<iostream> #include<cmath> using namespace std; long long M,i; #define LL long long int _work(LL a,LL n) { LL ans=1; while(n){ if(n&1){ ans=(ans*a)%M; n--; }…
题意: 思路: 对于每个幂次方,将幂指数的二进制形式表示,从右到左移位,每次底数自乘,循环内每步取模. #include <cstdio> typedef long long LL; LL Ksm(LL a, LL b, LL p) { LL ans = 1; while(b) { if(b & 1) { ans = (ans * a) % p; } a = (a * a) % p; b >>= 1; } return ans; } int main() { LL p, a…
Rightmost Digit Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 69614 Accepted Submission(s): 25945 Problem Description Given a positive integer N, you should output the most right digit of N…
链接: https://vjudge.net/problem/LightOJ-1282 题意: You are given two integers: n and k, your task is to find the most significant three digits, and least significant three digits of nk. 思路: 后三位快速幂取余,考虑前三位. \(n^k\)可以表示为\(a*10^m\)即使用科学计数法. 对两边取对数得到\(k*log…
Raising Modulo Numbers Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 5500 Accepted: 3185 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…