题目描述
include [HTML_REMOVED]
include [HTML_REMOVED]
using namespace std;
const int N = 1e6 + 10;
vector[HTML_REMOVED] add(vector[HTML_REMOVED] &A, vector[HTML_REMOVED] &B) {
vector[HTML_REMOVED] C;
int t = 0;
for (int i = 0; i < A.size() || i < B.size(); i++) {
if (i < A.size()) t += A[i];
if (i < B.size()) t += B[i];
C.push_back(t % 10);
t = t / 10;
}
if (t) C.push_back(1);
return C;
}
int main() {
string a, b;
vector[HTML_REMOVED] A, B, res;
cin >> a >> b;
for (int i = a.size() - 1; i >= 0; i–) A.push_back(a[i] - ‘0’);
for (int i = b.size() - 1; i >= 0; i–) B.push_back(b[i] - ‘0’);
res = add(A, B);
for (int i = res.size() - 1; i >= 0; i--) cout << res[i];
cout << endl;
return 0;
}
样例
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算法1
(暴力枚举) $O(n^2)$
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时间复杂度
参考文献
C++ 代码
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算法2
(暴力枚举) $O(n^2)$
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时间复杂度
参考文献
C++ 代码
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