算法导论——2.1-4

最近还找到一种方法：

#include<iostream>

using namespace std;

int main()
{
size_t n;
cin >> n;
int a[4], b[4],c[5];
cout << "input a[]:" << endl;
for (int i = 0; i < n; ++i)
cin >> a[i];
cout << "input b[]:" << endl;
for (int i = 0; i < n; ++i)
cin >> b[i];
for (int i = 0; i < n; ++i)
c[i + 1] = a[i] + b[i];
c[0] = 0;
for (int *p = end(c); p >= begin(c); --p)
{
if (*p == 2)
{
*(p - 1) = *(p - 1) + 1;
*p = 0;
}
}
for (int *p = begin(c); p < end(c); ++p)
cout << *p << " ";
cout <<endl;
return 0;
}不过对数组长度有限制，输入的n必须和数组的初始设定的长度一样，否则数组里面的垃圾数据会影响结果。