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…

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…

题目:http://poj.org/problem?id=1995 题目解析:求(A1B1+A2B2+ ... +AHBH)mod M. 大水题. #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> using namespace std; int n,mod,sum; int main() { ],b[…

快速幂取模 #include<cstdio> int mod_exp(int a, int b, int c) { int res, t; res = % c; t = a % c; while (b) { ) res = res * t % c; t = t * t % c; b >>= ; } return res; } int main() { int T; scanf("%d",&T); while(T--) { int m,h; scanf(&…

传送门:http://poj.org/problem?id=1845 大致题意: 求A^B的所有约数(即因子)之和,并对其取模 9901再输出. 解题基础: 1) 整数的唯一分解定理: 任意正整数都有且只有一种方式写出其素因子的乘积表达式. ,其中为素数 2) 约数和公式: 对于已经分解的整数,A的所有因子之和为 3) 同余模公式: (a+b)%m=(a%m+b%m)%m (a*b)%m=(a%m*b%m)%m 1: 对A进行素因子分解 这里如果先进行筛50000内的素数会爆空间,只能用最朴素的…

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…

Raising Modulo Numbers Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5477   Accepted: 3173 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…

POJ1995 Raising Modulo Numbers 计算(A1B1+A2B2+ ... +AHBH)mod M. 快速幂,套模板 /* * Created: 2016年03月30日 23时01分45秒 星期三 * Author: Akrusher * */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #…

Raising Modulo Numbers Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6347   Accepted: 3740 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…

题目链接:http://poj.org/problem?id=1995 解题思路:用整数快速幂算法算出每一个 Ai^Bi,然后依次相加取模即可. #include<stdio.h> long long quick_mod(long long a,long long b,long long c) { long long ans=1; while(b) { if(b&1) { ans=ans*a%c; } b>>=1; a=a*a%c; } return ans; } int…