典型的会一题吃所有
#include<iostream>
#define N 100010
using namespace std;
int a[N], b[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 res1[N],res2[N],res[N];
inline int Count(int x[],int y[],int len1,int len2)
{ //高精度加法板子，返回结果数组长度并将结果存在res数组中
int c[N] = {0},len = max(len1,len2);
for (int i = 0; i <= len; i++)
{
res[i] = x[i] + y[i] + c[i];
if(res[i] > 9)
{
c[i+1] = res[i] / 10;
res[i] %= 10;
}
}
while (res[len] != 0)
len++;
return len;
}
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 = Count(res1,res2,len1,len2);
for (int i = len-1; i >= 0; i--)
cout << res[i]; //反向输出结果
return 0;
}