POJ1995 Raising Modulo Numbers（快速幂模板）

最新推荐文章于 2022-08-05 21:11:12 发布

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最新推荐文章于 2022-08-05 21:11:12 发布

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分类专栏：题解————题解模板——模板

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本文介绍了一种名为快速幂的游戏，并详细解析了快速幂算法的基本原理及其应用。文章讲解了如何利用快速幂算法来解决特定数学游戏的问题，通过递推求解乘积项，实现了高效的计算。

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题目

描述
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 AiBi 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.
输入
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.
输出
For each assingnement there is the only one line of output. On this line, there is a number, the result of expression
(A1B1+A2B2+ … +AHBH)mod M.
样例输入
3
16
4
2 3
3 4
4 5
5 6
36123
1
2374859 3029382
17
1
3 18132
样例输出
2
13195
13

题解

快速幂模板题
那么我们来讲一下快速幂吧
对于任何一个数 $b$ 在二进制下有 $k$ 位，第 $i$ 位的数字为 $c_i$ 那么: $b=ck−1∗2k−1+ck−2∗2k−2+⋯+c0∗20b=c_{k-1}*2^{k-1}+c_{k-2}*2^{k-2}+\cdots+c_0*2^0$ 于是: $ab=ack−1∗2k−1∗ack−2∗2k−2∗⋯∗ac0∗20a^b=a^{c_{k-1}*2^{k-1}}*a^{c_{k-2}*2^{k-2}}*\cdots*a^{c_0*2^0}$
因为 $k=⌈log2(b+1)⌉k=\left \lceil log_2(b+1) \right \rceil$ ，所以上式的乘积项不多于 $⌈log2(b+1)⌉\left \lceil log_2(b+1) \right \rceil$ 个，又因为: $a^{2^{i}} = (a^{2^{i-1}})^2$
所以我们很容易通过 $k$ 次递推求出每个乘积项，当 $c_i=1$ 时，把该乘积项累加到答案中。 $b$ & $1$ 可以取出 $b$ 在二进制下的最低位，而 $b > > 1$ 可以舍弃 $b$ 在二进制下的最后一位，将二者结合就可以遍历 $b$ 在二进制下的所有数位 $c_i$ ，整个算法的时间复杂度为 $O(log_{2}b)$

code

LL t, m, h, x, y, sum;
LL s[maxn];

LL quick_power(LL a, LL b, LL p) {
	LL ans = 1 % p;
	for (; b; b >>= 1) {
		if (b & 1) ans = ans * a % p;
		a = a * a % p;
	}
	return ans;
}

int main() {
	read(t);
	while (t--) {
		sum = 0;
		read(m); read(h);
		for (int i = 1; i <= h; ++i) {
			read(x), read(y);
			s[i] = quick_power(x, y, m);
			s[i] %= m;
		}
		for (int i = 1; i <= h; ++i)
			sum = (sum + s[i]) % m;
		write(sum), putchar('\n');
	}
	return 0;
}