以C为模的大数的幂 [英] Power for big numbers modulo m in C

问题描述

我正在使用以下函数来计算以m为模的大数的幂，其中m是任何整数，即（a ^ b）％m

I am using the following function to compute powers of big numbers modulo m, where m is any integer,i.e. (a^b)%m

long long power(long long x, long long y, long long p)
{
    long long res = 1;      // Initialize result

    x = x % p;  // Update x if it is more than or
                // equal to p

    while (y > 0)
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res*x) % p;

        // y must be even now
        y = y>>1; // y = y/2
        x = (x*x) % p;
    }
    return res;
}

但是，对于某些数字，即使此功能也不起作用。例如，如果我呼叫

But, for some numbers even this function is not working. For example, if I call

 power(1000000000000,9897,52718071807);

我得到一个负数作为输出。发生这种情况是由于以下原因：
幂函数中有一行：

I get a negative number as the output. It happens because of the following reason: There is a line in the power function:

  x = (x*x) % p;

当x大时，假设x = 46175307575，执行x =（ x * x）％p变为负数。我不明白为什么会这样。即使（x * x）的值超过了long long int的上限，我也不会将其值存储在任何地方，而只是存储（x * x）％p，其值应介于0到p之间。另外，由于p没有跨过长距离，x会如何交叉呢？请告诉我为什么会出现此问题以及如何解决此问题。

When x is big, let's say x=46175307575, the value stored in x after executing x=(x * x)%p becomes negative. I don't understand why it happens. Even if the value of (x * x) crosses the upper range of long long int, I am not storing its value anywhere, I am just storing (x*x)%p whose value should lie between 0 to p. Also, since p doesn't crosses the long long range, how does x cross it? Please tell me why this problem occurs and how to solve this.

推荐答案

在 GeeksForGeeks 是以下功能：

// Returns (a * b) % mod
long long moduloMultiplication(long long a,
                               long long b,
                               long long mod)
{
    long long res = 0;  // Initialize result

    // Update a if it is more than
    // or equal to mod
    a %= mod;

    while (b)
    {
        // If b is odd, add a with result
        if (b & 1)
            res = (res + a) % mod;

        // Here we assume that doing 2*a
        // doesn't cause overflow
        a = (2 * a) % mod;

        b >>= 1;  // b = b / 2
    }

    return res;
}

使用它代替

x = (x*x) % p;

即，

x = moduloMultiplication(x, x, p);

，而不是

res = (res*x) % p

即，

res = moduloMultiplication(res, x, p);

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