大數相乘

使用分治法實現大數相乘，不要求兩數長度一致。時間複雜度爲T(N)=O(N^1.59).

#include <iostream>

#include <math.h>

using namespace std;

int SIGN(long A);

long CalculateUnSame(long X, long Y, int xn, int yn);

int main(int argc, char *argv[])

{

    cout<<"please input two numbers: \nX="; //若是用getline獲取string，則應用stringstream轉爲long.

    long X = 0;

    cin>>X;

    cout<<"Y=";

    long Y = 0;

    cin>>Y;

    int xn = 1;

    long x1 = X;

    while( x1 / 10 > 0) //求長度

    {

        xn++;

        x1 /= 10;

    }

    int yn = 1;

    long y1 = Y;

    while( y1 / 10 > 0)

    {

        yn++;

        y1 /= 10;

    }

    long sum = CalculateUnSame(X, Y, xn,yn);

    cout<<"normal mult_result X * Y = "<<X<<" * "<<Y<<" = "<<X * Y<<endl;

    cout<<"divide and conqure X * Y = "<<X<<" * "<<Y<<" = "<<sum<<endl;

    return 0;

}

int SIGN(long A)

{

    return A > 0 ? 1 : -1;

}

long CalculateUnSame(long X, long Y, int xn, int yn)

{

    int sign = SIGN(X) * SIGN(Y);

    X = abs(X);

    Y = abs(Y);

    if (X == 0 || Y == 0)

        return 0;

    else if (xn == 1 || yn == 1)

        return  sign * X * Y;

    else

    {

        int xn0 = xn / 2, yn0 = yn / 2;

        int xn1 = xn - xn0, yn1 = yn - yn0;

        long A = (long)(X / pow(10, xn0));

        long B = (long)(X % (long)pow(10, xn0));

        long C = (long)(Y / pow(10, yn0));

        long D = (long)(Y % (long)pow(10, yn0));

        long AC = CalculateUnSame(A, C, xn1, yn1);

        long BD = CalculateUnSame(B, D, xn0, yn0);

        long ABCD = CalculateUnSame((long)(A * pow(10, xn0) - B), (long)(D - C * pow(10, yn0)), xn1, yn1);

        return (long) sign * (2 * AC * pow(10, (xn0 + yn0)) + ABCD + 2 * BD);

    }

}

上述代碼中數據類型定義爲long，而在limits.h中可以看到：long型數最大值爲2147483647，最小值爲-2147483648

開根號得46340，五位數。所以上面代碼計算結果大於 LONG_MAX時就會溢出。

將數據類型改爲long long可以計算的結果爲：最大值9223372036854775807，最小值-9223372036854775808

根號9223372036854775807得到3037000500，10位數。

不管是long還是long long都有長度限制。所以本算法並不能真正做到大數相乘。進一步的改進是將數據都用string來傳遞。