# [POJ 1995] Raising Modulo Numbers【快速幂】

## Problem:

Time Limit: 1000MS |
Memory Limit: 30000K |

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 this market segment was so far underestimated and that there is lack of such games. This kind of game was thus included into the KOKODáKH. The rules follow:

Each player chooses two numbers Ai and Bi and writes them on a slip of paper. Others cannot see the numbers. In a given moment all players show their numbers to the others. The goal is to determine the sum of all expressions Ai^{Bi} from all players including oneself and determine the remainder after division by a given number M. The winner is the one who first determines the correct result. According to the players' experience it is possible to increase the difficulty by choosing higher numbers.

You should write a program that calculates the result and is able to find out who won the game.

Input

Output

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

`(A _{1}^{B1}+A_{2}^{B2}+ ... +A_{H}^{BH})mod M`.

Sample Input

3 16 4 2 3 3 4 4 5 5 6 36123 1 2374859 3029382 17 1 3 18132

Sample Output

2 13195 13

Source

## Solution:

一道大水题。

最基本的快速幂计算。

## Code: O(ZHlogH) [164K, 125MS]

#include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; typedef long long ll; int Z, H; ll M, A, B; inline ll fastpow(ll bas, ll ex, ll md){ ll iden = bas, res = 1; while(ex){ if(ex & 1) res = res * iden % md; ex >>= 1, iden = iden * iden % md; } return res; } int main(){ scanf("%d", &Z); while(Z--){ scanf("%lld%d", &M, &H); ll res = 0; while(H--){ scanf("%lld%lld", &A, &B); res += fastpow(A, B, M); if(res >= M) res -= M; } printf("%lld\n", res); } return 0; }

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