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。