Poj1995 Raising Modulo Numbers(快速幂)

本文介绍了一种数学游戏的编程解决方案,该游戏要求玩家计算特定表达式的模运算结果。文章详细解释了利用二进制表示和快速幂算法来高效解决这一问题的方法,并提供了完整的C++代码示例。

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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+A2^B2+ … +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
HINT
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=ck2k1+ck12k2+...+c020b=ck2k−1+ck−12k−2+...+c020

于是:
ab=ack12k1ack22k2...ac020ab=ack−1∗2k−1∗ack−2∗2k−2∗...∗ac0∗20

因为k=log2(b+1)k=⌈log2(b+1)⌉(其中 ⌈⌉ 表示向上取整),所以上式乘积项的数量不多于log2(b+1)⌈log2(b+1)⌉ 个。又因为a2i=(a2i1)2a2i=(a2i−1)2,所以我们很容易通过k次递归求出每个乘积项,当ci=1ci=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cici。整个算法的时间复杂度为O(log2b)O(log2b)

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;
}
### 关于POJ 1995问题的快速幂C++实现 对于POJ 1995问题,其核心在于通过矩阵快速幂算法高效解决大规模数据下的指数运算。以下是基于引用材料中的相关内容构建的一个完整的解决方案。 #### 矩阵快速幂的核心逻辑 矩阵快速幂是一种高效的计算方式,在处理线性递推关系时尤为有效。例如斐波那契数列可以通过构造特定的转移矩阵来加速计算[^4]。具体来说,给定一个初始状态向量和一个转移矩阵,经过若干次幂运算后可获得目标状态。 以下是一个通用的矩阵快速幂模板: ```cpp #include <iostream> using namespace std; const int N = 2; // 定义矩阵大小 struct Matrix { long long m[N][N]; }; // 矩阵乘法函数 Matrix multiply(const Matrix& a, const Matrix& b) { Matrix c; for (int i = 0; i < N; ++i) { for (int j = 0; j < N; ++j) { c.m[i][j] = 0; for (int k = 0; k < N; ++k) { c.m[i][j] += a.m[i][k] * b.m[k][j]; c.m[i][j] %= 10000; // 取模操作 } } } return c; } // 快速幂函数 Matrix fastPower(Matrix base, long long exp) { Matrix result; for (int i = 0; i < N; ++i) { for (int j = 0; j < N; ++j) { result.m[i][j] = (i == j); } } while (exp > 0) { if (exp % 2 == 1) { result = multiply(result, base); } base = multiply(base, base); exp /= 2; } return result; } int main() { long long n; cin >> n; // 初始化转移矩阵 Matrix trans; trans.m[0][0] = 0; trans.m[0][1] = 1; trans.m[1][0] = 1; trans.m[1][1] = 1; // 计算结果矩阵 if (n == 0 || n == 1) { cout << 1 << endl; } else { Matrix res = fastPower(trans, n - 1); // 初始状态向量 long long fib_prev = 1; long long fib_curr = 1; // 输出结果 cout << (res.m[0][0] * fib_prev + res.m[0][1] * fib_curr) % 10000 << endl; } return 0; } ``` 此代码实现了针对斐波那契数列的大规模项求解功能,并采用了取模`%10000`的操作以满足题目需求。其中的关键部分包括矩阵乘法、快速幂以及状态转移的设计[^3]。 #### 特殊注意点 在实际提交过程中需要注意以下几个方面: - **大数组定义**:如果涉及更大的矩阵或者更复杂的动态规划表,则需特别留意内存分配的位置及其范围限制[^2]。 - **时间复杂度控制**:尽管快速幂本身具有较低的时间复杂度O(log n),但在极端情况下仍需验证是否存在进一步优化空间。 - **边界条件处理**:如输入为较小数值时直接返回已知答案而非进入循环计算流程。
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