注意点都在注释里
#include<iostream>
#define N 100010
using namespace std;
int a[N], b[N];
int res1[N], res2[N];
inline int Mul(int a[], int lena, int b[], int lenb)
{ //和前几题一样的高精度板子
int c[N] = { 0 };
int lenc = lena + lenb;
for (int i = 0; i < lena; i++)
{
for (int j = 0; j < lenb; j++)
{
c[i + j] += a[i] * b[j];
c[i + j + 1] += c[i + j] / 10;
c[i + j] %= 10;
}
}
if (c[lenc - 1] == 0 && lenc > 0)
lenc--;
for (int i = 0; i < lenc; i++)
a[i] = c[i];
return lenc;
}
inline int Input(int x[], int y[], int m, int n, int res[])
{ //封装输入函数并返回高精度乘法处理后的数组长度，将结果存在res1/2数组中
int len, lena;
int i = 0;
while (m != 0)
{
x[i] = m % 10;
y[i] = m % 10;
m /= 10;
i++;
}
len = lena = i;
for (int i = 1; i < n; i++)
lena = Mul(a, lena, b, len);
for (int i = 0; i < lena; i++)
res[i] = a[i];
return lena;
}
int main()
{
int x1, n1, x2, n2;
int len1, len2, len;
cin >> x1 >> n1 >> x2 >> n2;
len1 = Input(a, b, x1, n1, res1);
len2 = Input(a, b, x2, n2, res2);
len = Mul(res1, len1, res2, len2);
//这里注意计算的最终结果回传到了res1数组中！
for (int i = len - 1; i >= 0; i--)
cout << res1[i]; //反向输出结果
return 0;
}