可能是新解法+高精求助

ShwStone @ 2021-10-18 20:43:31

#include <iostream>
using namespace std;

int main() {
    int a, b;
    cin >> a >> b;
    while (b) {
        a ^= b;
        b = ((a ^ b) & b) << 1;
    }
    cout << a << endl;
    return 0;
}

by whitenessP @ 2021-10-18 20:45:03

应该 a+b 更快

by int64 @ 2021-10-18 20:45:35

@ShwStone bitset 我之前实现了一下，会 TLE

by int64 @ 2021-10-18 20:46:13

显然也有可能我的解法不好。

record

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<cstring>
#include<iomanip>
#include<vector>
#include<bitset>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);

using namespace std;

#define inf 0x3f3f3f3f
#define int long long
#define endl '\n'

string to2(int n) {
    string ret;
    while (n) {
        ret += (n % 2 + '0');
        n /= 2;
    }
    reverse(ret.begin(), ret.end());
    return ret;
}

signed main() {
    IOS;
    int a, b;
    cin >> a >> b;
    if (b > a) {
        swap(a, b);
    }
    bitset<70> res(to2(a));
    while (b--) {
        for (int i = 0;i < 64;i++) {
            if (res[i] == 1) {
                res[i] = 0;
            }
            else {
                res[i] = 1;
                break;
            }
        }
    }
    cout << res.to_ullong() << endl;
    return 0;
}

//Fununugugu Code

by x义x @ 2021-10-18 20:46:44

草，可以，很有想法

by WYXkk @ 2021-10-18 20:50:27

@ShwStone 你这个做法相当于把加法的底层实现用 cpp 写出来了

by UnyieldingTrilobite @ 2021-10-18 20:56:50

@ShwStone 抱歉，这并不是新解法

by ShwStone @ 2021-10-18 20:58:49

@int64 我的意思是完全模拟计算机，就像这样：

#include <bits/stdc++.h>
using namespace std;

bitset<70> operator + (bitset<70> a, bitset<70> b) {
    while (b.count() > 0) {
        a ^= b;
        b = ((a ^ b) & b) << 1;
    }
    return a;
}

int main() {
    int a, b;
    bitset<70> x, y, z;
    cin >> a >> b;
    if (a >= 0) x = a;
    else {
        x = -a;
        x = ~x;
        x = x + bitset<70>(1);
    }
    if (b >= 0) y = b;
    else {
        y = -b;
        y = ~y;
        y = y + bitset<70>(1);
    }
    z = x + y;
    if (z[69] == 1) {
        cout << '-';
        z = ~z;
        z = z + bitset<70>(1);
    }
    cout << z.to_ulong() << endl;
    return 0;
}

AC了。

record:

by ShwStone @ 2021-10-18 21:00:12

@UnyieldingTrilobite 确实，只是模拟计算机

by NetherDevil @ 2021-10-18 21:00:57

@ShwStone 不知道 x86 有没有 MIPS 类似的 add 溢出检测，但是估计即使有洛谷也不会让用 asm 关键字

by int64 @ 2021-10-18 21:05:11

草了，看不懂。