Poj1995 Raising Modulo Numbers（快速幂）

Description
People are different. Some secretly read magazines full of interesting girls' pictures, others creat
e 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 i
s 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 numbe
rs. In a given moment all players show their numbers to the others. The goal is to determine the sum
of all expressions AiBi from all players including oneself and determine the remainder after divisi
on 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+A2B2+ ... +AH^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

题意：求a的b次方对p取模的值，其中1<=a,b,p<=1000000000；

题解：根据数学常识，每一个正整数可以唯一表示为若干指数不重复的2的次幂的和。

也就是说，如果b在二进制表示下有k位，其中第i（0<=i<=k）位的数字是ci，那么：

\[b=c_k 2^{k-1}+c_{k-1} 2^{k-2}+ ... +c_0 2^0 \]

于是：$$ a^b=a{c_{k-1}2^{k-1}}*a{c_{k-2}2^{k-2}}*...*a{c_0*2^0} $$
因为$ k=\lceil log_2(b+1) \rceil $（其中 \(\lceil \rceil\) 表示向上取整），所以上式乘积项的数量不多于\(\lceil log_2(b+1) \rceil\) 个。又因为\(a^{2^i}=(a^{2{i-1}})^2\)，所以我们很容易通过k次递归求出每个乘积项，当\(c_i=1\)时，把该乘积项累积到答案中。b&1 运算可以取出b在二进制表示下的最低位，而b>>1运算可以舍去最低位，在递推的过程中将二者结合，就可以遍历b在二进制表示下的所有数位\(c_i\)。整个算法的时间复杂度为\(O(log_2b)\)。

代码如下：

#include<bits/stdc++.h>
using namespace std;
int read()
{
	int x=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int power(int a,int b,int p)
{
	int ans=1%p;
	while(b)
	{
		if(b&1)ans=(long long)ans*a%p;
		a=(long long)a*a%p;
		b>>=1;
	}
	return ans;
}
int main()
{
	int T,k,a,b,n;
	T=read();
	while(T--)
	{
		int ans=0;
		k=read();n=read();
		for(int i=1;i<=n;i++)
		{
			a=read();b=read();
			ans=(ans+power(a,b,k))%k;
		}
		printf("%d\n",ans);
	}
	return 0;
}