Poj1995--Raising Modulo Numbers（快速幂）

题目: http://poj.org/problem?id=1995

Input

The input consists of Z assignments. The number of them is given by the single positive integer Z appearing on the first line of input. Then the assignements follow. Each assignement begins with line containing an integer M (1 <= M <= 45000). The sum will be divided by this number. Next line contains number of players H (1 <= H <= 45000). Next exactly H lines follow. On each line, there are exactly two numbers Ai and Bi separated by space. Both numbers cannot be equal zero at the same time.

Output

For each assingnement there is the only one line of output. On this line, there is a number, the result of expression

(A₁^B₁+A₂^B₂+ ... +A_H^B_H)mod M.

快速幂，裸题：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
long long Deal(long long a, long long b, int mod)
{long long ans = 1;while(b){if(b & 1)ans = ans*a%mod;b /= 2;a = (a*a)%mod;    }    return ans;
}
int main()
{int t;scanf("%d", &t);while(t--){int mod, h;long long a, b, sum = 0;scanf("%d %d", &mod, &h);for(int i = 0; i < h; i++){scanf("%lld %lld", &a, &b);sum += Deal(a, b, mod) % mod; }printf("%lld\n", sum%mod);}return 0;
}

转载于:https://www.cnblogs.com/soTired/p/4755988.html

Poj1995--Raising Modulo Numbers（快速幂）相关推荐

poj 1995 Raising Modulo Numbers 二分快速幂
题意:给定n对Ai,Bi,求所有Ai的Bi次方之和对M取模的结果: 思路:二分法求快速幂: #include<cstdio> #include<cstring> #includ ...
【poj1995】Raising Modulo Numbers
problem T组数据,每组包含n对ai,bi和一个p. 每组输出一个答案,∑ni=1aibi%p∑i=1naibi%p \sum_{i=1}^n ai^{bi} \%p. solution 快速幂 ...
Raising Modulo Numbers
http://poj.org/problem?id=1995 题解:快速幂 /* *@Author: STZG *@Language: C++ */ //#include <bits/stdc+ ...
【POJ - 1995】Raising Modulo Numbers（裸的快速幂）
题干: People are different. Some secretly read magazines full of interesting girls' pictures, others c ...
c语言的幂乘积表达式,POJ 1845 Sumdiv [素数分解快速幂取模二分求和等比数列]
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 ...
[矩阵乘法/快速幂专题]Arc of Dream，Recursive sequence，233 Matrix，Training little cats
矩阵快速幂习题复习矩阵乘法及快速幂模板乘法模板快速幂模板 T1:Arc of Dream 题目题解 code T2:Recursive sequence 题目题解 code T3:233 M ...
`Computer-Algorithm` 数论基础知识 (同余,取模,快速幂,质数,互质,约数,质因子)
catalog 同余取模快速幂质数互质约数质因子 @Delimiter(旧解释) 经验谈两数之差也整除加一的特殊性取模累加的周期性取模的唯一集合取模下的四则运算除法的不可约性 ...
矩阵快速幂2Jzzhu and Sequences Recurrences Contemplation! Algebra Reading comprehension
来!接着写题解,立个flag:矩阵快速幂题解不写完今晚不睡觉! 此题解是接着矩阵快速幂1写的,此篇有哪里写不清楚的地方导致看不懂的可以先看1:https://blog.csdn.net/qq_4536 ...
等比数列求和（快速幂 + 逆元）
求一个等比数例之和, 并让他对一个数取模. 用到等比数列求和公式, 快速幂, 逆元. 不会证明, 下面给出代码. #include <stdio.h> #include <strin ...

Poj1995--Raising Modulo Numbers（快速幂）

Poj1995--Raising Modulo Numbers（快速幂）相关推荐

最新文章

热门文章