uva 10469번 To Carry or not to Carry

uva : https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=16&page=show_problem&problem=1410

 

10진수 -> 2진수로 변경 후

carry 값을 0으로 만들어주면 끝나는 매우 쉬운 문제

 

#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#include <set>
#include <cmath>
#include <limits>
#include <cstring>
#include <string>
using namespace std;
 
int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
 
    long long n1, n2;
    int a[40], b[40];
    while (1) {
        cin >> n1 >> n2;
        if (cin.eof()) break;
 
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
 
        // 10진수 -> 2진수
        string astr = "", bstr = "";
        while (n1) {
            if (n1 == 1) {
                astr += '1';
                break;
            }
            astr += (n1 % 2) + '0';
            n1 /= 2;
        }
        while (n2) {
            if (n2 == 1) {
                bstr += '1';
                break;
            }
            bstr += (n2 % 2) + '0';
            n2 /= 2;
        }
        for (int i = astr.size() - 1; i >= 0; i--) a[i] = astr[i] == '1' ? 1 : 0;
        for (int i = bstr.size() - 1; i >= 0; i--) b[i] = bstr[i] == '1' ? 1 : 0;
 
        int len = max(astr.size(), bstr.size());
        long long ans = 0;
        for (int i = 0; i < len; i++) {
            if (a[i] && b[i]) continue;
            if (a[i] || b[i]) ans += pow(2, i);
        }
        cout << ans << "\n";
    }
 
    return 0;
}