Raising Modulo Numbers

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最新推荐文章于 2024-08-17 03:37:10 发布

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原文链接：http://www.cnblogs.com/heqinghui/archive/2012/07/24/2605894.html

本文介绍了一种结合数学与竞争的游戏——模数运算游戏，玩家通过选择两个数字并进行运算，最终计算出所有表达式的结果对特定整数的余数。文章详细解释了游戏规则、输入输出格式，并提供了实现该游戏所需算法的C++代码示例。重点在于理解模运算的性质及其在编程中的应用，旨在提高参与者对数学问题解决能力和编程技巧。

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/*Raising Modulo Numbers

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3555		Accepted: 1964

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

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.

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
*/
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cmath>
using namespace std;
int powmod (int a,int b,int n)
{
 if(b==0)return 1;
 int k=1;
 while(b>=1)
 {
  if(b%2!=0)k=a*k%n;
  a=((a%n)*(a%n))%n;//算a^2 mod n因为计算机可以保存中间结果，减少计算时间。
  b/=2;
 }
 return k;
}
int main()
{
 int a,b,t,n,k,sum,i;
 scanf("%d",&t);
 while(t--)
 {
  sum=0;
  scanf("%d%d",&n,&k);
  for(i=0;i<k;i++)
  {
   scanf("%d%d",&a,&b);
   sum+=powmod(a,b,n);
   sum%=n;
  }
  printf("%d\n",sum);
 }
 return 0;
}