http://poj.org/problem?id=1995 简单的快速幂问题 要注意num每次加过以后也要取余,否则会出问题 #include<iostream> #include<cstdio> using namespace std; typedef long long ll; ll mod_pow(ll x,ll n,ll mod) { ll res=; ) { ) res=res*x%mod; x=x*x%mod; n>>=; } return res; }…

Raising Modulo Numbers Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5934   Accepted: 3461 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…

题意:给出A1,…,AH,B1,…,BH以及M,求(A1^B1+A2^B2+ … +AH^BH)mod M. 思路:快速幂 实例 3^11  11=2^0+2^1+2^3    => 3^1*3^2*3^8=3^11 实现代码: int solve(int a,int b) { int ans=1;          while(b){ if(b&1)  ans=ans*a;   a=a*a;  b>>=1;} } 解释一下代码:b&1即二进制表达式的最后一位,  11二…

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…

Sumdiv Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1845 Appoint description:   System Crawler  (2015-05-27) Description Consider two natural numbers A and B. Let S be the sum of all natural…

题意:       给你一个无向图,然后给了一个起点s和终点e,然后问从s到e的最短路是多少,中途有一个限制,那就是必须走k条边,路径可以反复走. 思路:       感觉很赞的一个题目,据说证明是什么国家队集训队论文什么的,自己没去看那个论文,就说下我自己的理解吧,对于这个题目,我们首先分析下Floyd,那个算法的过程中是在更新的dis[i][j]上再更新,再更新...,是想一下,我们每次都把更新的结果存下来,就是每次答案数组初始化全是INF,然后用当前的dis数组和原始的map来更新,那么更…

http://poj.org/problem?id=3641 练手用,结果念题不清,以为是奇偶数WA了一发 #include<iostream> #include<cstdio> #include<cmath> using namespace std; typedef long long ll; bool judge_prime(ll k) { ll i; ll u=int(sqrt(k*1.0)); ;i<=u;i++) { ) ; } ; } ll mod_p…

#include <cstring> #include <cstdio> #include <iostream> #include <cmath> #include <algorithm> using namespace std; #define LL long long LL p,res,a; bool judge_prime(LL k) { LL i; LL u=int(sqrt(k*1.0)); ;i<=u;i++) { ) ; }…

Description Fermat'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 not very many) non-prime values of p, known as base-a ps…

嗯... 题目链接:http://poj.org/problem?id=1995 快速幂模板... AC代码: #include<cstdio> #include<iostream> using namespace std; int main(){ ; scanf("%lld", &N); while(N--){ scanf("%lld%lld", &M, &n); sum = ; ; i <= n; i++){…