E. Vlad and a Pair of Numbers

合集 - codeforces(31)

1.B. Rudolf and 1212024-03-162.C. Rudolf and the Ugly String2024-03-163.A. Rudolf and the Ticket2024-03-164.D. Rudolf and the Ball Game2024-03-185.G. Rudolf and Subway2024-04-066.C. Choose the Different Ones!2024-02-097.B. Following the String2024-02-098.E. Klever Permutation2024-02-099.D. Find the Different Ones!2024-02-0910.B. YetnotherrokenKeoard2023-12-1011.C. Removal of Unattractive Pairs2023-12-1012.D. Jumping Through Segments2023-12-1113.E. Good Triples2023-12-1314.F. Shift and Reverse2023-12-1315.C. Nikita and LCM2024-05-2916.B. Prefiquence2024-05-03

17.E. Vlad and a Pair of Numbers2024-04-07

18.D. XOR Construction2024-03-2719.C. Non-coprime Split2024-01-2720.D. Divide and Equalize2024-06-0221.E. Block Sequence2024-06-0222.D. In Love2024-06-1123.C. Theofanis' Nightmare2024-06-2224.D. Yet Another Monster Fight2024-06-2225.C. Count Triangles2024-07-1126.E. Count Paths2024-07-1327.Collecting Numbers II2024-07-3128.C. Even Subarrays2024-08-0729.C. Perform Operations to Maximize Score2024-08-2030.F. Color Rows and Columns2024-08-1631.C. To Become Max2024-09-10

收起

题解

首先，我们知道异或运算是无进位相加，那么a^b=x我们不妨先让a=x，b=0；而a，b其余二进制位上要么同为0，要么同为1。接下来，根据题意a+b=2x，我们可知我们同时为a，b加上x/2。此时再判断a^b是否等于x即可。

code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    int t;
    cin>>t;
    while (t--){
        ll x;
        ll a,b;
        cin>>x;
        a=x/2;
        b=x+x-a;
        if ((a^b)==x) cout<<a<<" "<<b<<endl;
        else cout<<-1<<endl;
    }
    return 0;
}