算法-大数乘法（16进制）（C语言实现）

由来

由于工作中用到，自己在网上找了一些大数乘法源码，都没有符合我要求的。所以自己造轮子，实现了一个。

分析过程

仿照我们小学学过的10进制乘法算式，来实现的大数乘法。

先举一个例子：234 * 567



这是我们最普通的乘法算式，对于程序实现起来还有点难，因此我们再继续计算分析：

（注：图中的小数字，不是次方，只是进位的数）



这个计算过程跟上图的区别就是加法跟乘法分开来计算，我在做乘法部分的时候，只是单纯的计算乘法，没有做加法，把进位的数标记出来，最后把在同一列的数字统统加起来，该进位的进位，最后起算出来结果。

写到这里大概就清晰了，我们先列个数组。

//用伪代码数组表示数：234，567
a[3] = {2, 3, 4}
b[3] = {5, 6, 7}

下图为数组中表示形式

程序讲解

以上讲解为分析过程，我用我们常用的10进制作为分析，在下面源码中用到了0x100进制，其实原理是一样的，一个unsigned char类型变量，正好最大可以支持到0xFF。

程序源码

以下是大数乘法编译过得程序

#include <stdio.h>

typedef unsigned char u8;
typedef unsigned int u32;

#define N 32
#define CARRYNUM 0x100

void BigMul(u8* res_high, u8* res_low, u8 *a, u8 *b)
{
int ai, bi, ri;
u32 tmp;
u32 res[N*2] = {0};

//乘法计算核心部分
for(ai = N-1; ai >= 0; ai--)    //a从后往前，低位往高位乘起来
{
for(bi = N-1; bi >=0; bi--)    //b从后往前，低位往高位乘起来
{
tmp = a[ai] * b[bi];
res[ai+bi+1] += (tmp % CARRYNUM); //取低位，并累加所有相同下标
res[ai+bi] += (tmp / CARRYNUM);   //取高位，并累加所有相同下标
}
}
for(ri = (2*N)-1; ri > 0; ri--)
{
if(res[ri] > CARRYNUM-1)    //从res低位开始判断，若大于0x100，向高位进位
{
res[ri-1] += (res[ri]/CARRYNUM);    //低位 进位数加到 高位
res[ri] = res[ri]%CARRYNUM;         //低位数的进位去掉，保留低位数的低位
}
}

//输出
for(ri = 0; ri < 2*N; ri++)
{
if(ri < N)
res_high[ri] = res[ri];
else
res_low[ri-N] = res[ri];
}
return;
}

void print_num(u8* res_high, u8* res_low)
{
int i = 0;
for(i = 0; i < N; i++)
{
printf("%02x ", res_high[i]);
}

for(i = 0; i < N; i++)
{
printf("%02x ", res_low[i]);
}
}

void main()
{
//数组[0]数据高位, [31]数据低位
//u8 a
= {0xDE,0x2A,0x7C,0xF8,0x48,0x24,0xCF,0xA0,0xCE,0xD7,0x16,0xAD,0xF5,0xD4,0x7D,0xE9,0x3D,0xFA,0x31,0xEB,0x69,0x26,0xB7,0xF7,0xDF,0x69,0xBD,0xE0,0x1F,0x9E,0xDB,0x0D};
//u8 b
= {0x5A,0x1C,0x23,0x99,0x87,0x70,0x82,0xB8,0x12,0x50,0xCB,0x2A,0x7F,0x8B,0x74,0x53,0x5D,0x38,0x05,0xB5,0xB0,0xFE,0x8C,0x0E,0xF1,0xE7,0x73,0x15,0x47,0xD4,0x9A,0xA2};

//u8 a
= {0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00};
//u8 b
= {0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00};

//u8 a
= {0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11};
//u8 b
= {0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11};

//u8 a
= {0x00,0x00,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11};
//u8 b
= {0x00,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11};

u8 a
= {0x00,0x00,0x00,0xD1,0xB8,0xAA,0x3E,0x1B,0x32,0xDB,0xB9,0x5A,0x13,0xB3,0x52,0x79,0x10,0x22,0x1E,0x47,0x49,0x00,0x8D,0xF9,0xA3,0x40,0xE6,0x9D,0x68,0x38,0x2C,0x58};
u8 b
= {0x00,0x00,0x00,0x00,0x00,0xBF,0xB7,0x4C,0x0C,0xBF,0x09,0x13,0x1E,0x1E,0xEC,0x72,0x21,0x60,0xE5,0x5F,0xAD,0xCF,0x8E,0x50,0x50,0xE7,0xBA,0xF5,0xE9,0x6B,0x3D,0x5E};

u8 res_high
= {0};
u8 res_low
= {0};
int i = 0;

BigMul(res_high, res_low, a, b);

print_num(res_high, res_low);

}

以下是执行结果：

00 00 00 00 00 00 00 00 9d 0e f0 53 4c 63 04 72 6f 67 38 02 90 dd a1 d9 3a 25 0d df d2 4a d4 b4 7b 5a 3a 1c e7 9e db 99 6e 16 0d 00 2e b8 06 be bd 5c c0 38 37 8a 41 ff 21 d0 3f b0 47 f9 40 50 请按任意键继续…

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