记录大数相乘的一般思路,乘法只是正数的大数相乘,数组第0位代表数组长度。
z[1]=x[1]*y[1]
z[2]=x[1]*y[2] +x[2]*y[1]
z[3]=x[1]*y[3] +x[2]*y[2]+x[3]*y[1]
…
因此可得z[i]=x[i-j+1]*y[j], for j=1到 y[0], x[0]> i-j>=0
考虑到进为问题,得到如下程序
Init(t);
t[0] = x[0] + y[0] - 1;
for (i = 1; i <= t[0]; i++)
{
sum = carry;
carry = 0;
for (j = 1; j <= y[0]; j++)
{
if ((i - j) >= 0 && (i - j) < x[0])
{
mul = x[i - j + 1];
mul = (unsigned long long)mul * y[j];
carry = carry + mul / 0x100000000;
mul = mul & 0xffffffff;
sum = sum + mul;
}
}
carry = carry + sum / 0x100000000;
t[i] = (unsigned int)sum;
}
全部代码如下
/****************************************************************************************************
大数相乘
调用方式Mul(x,y,z)
返回值,z=x*y
*****************************************************************************************************/
void Mul_Long(unsigned int x[], unsigned long long y, unsigned int *z)
{
unsigned int t[34];
unsigned long long mul;
unsigned int carry = 0;
int i;
Init(t);
Mov_Big(x, t);
for (i = 1; i <= x[0]; i++)
{
mul = x[i];
mul = mul * y + carry;
t[i] = (unsigned int)mul;
carry = (unsigned int)(mul >> 32);
}
if (carry != 0)
{
t[0]++;
t[t[0]] = carry;
}
i = t[0];
while (t[i] == 0 && i > 1) //这里有改动i>1
{
t[0]--;
i--;
}
Mov_Big(t, z);
}
void Mul_Big(unsigned int x[], unsigned int y[], unsigned int *z)
{
unsigned int t[100];
unsigned long long sum, mul = 0, carry = 0;
unsigned int i, j;
if (y[0] == 1)
Mul_Long(x, y[1], z);
else
{
Init(t);
t[0] = x[0] + y[0] - 1;
for (i = 1; i <= t[0]; i++)
{
sum = carry;
carry = 0;
for (j = 1; j <= y[0]; j++)
{
if ((i - j) >= 0 && (i - j) < x[0])
{
mul = x[i - j + 1];
mul = (unsigned long long)mul * y[j];
carry = carry + mul / 0x100000000;
mul = mul & 0xffffffff;
sum = sum + mul;
}
}
carry = carry + sum / 0x100000000;
t[i] = (unsigned int)sum;
}
if (carry != 0)
{
t[0]++;
t[t[0]] = (unsigned int)carry;
}
i = t[0];
while (t[i] == 0 && i > 1)//这里有改动
{
t[0]--;
i--;
}
Mov_Big(t, z);
}
}